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\begin{center}
\vskip 1cm{\LARGE\bf 
Involutions on Generating Functions
}
\vskip 1cm
\large
Masanari Kida \\
Department of Mathematics \\
University of Electro-Communications \\
Chofu, Tokyo 182-8585  \\
Japan \\
\href{mailto:masanari.kida@gmail.com}{\tt masanari.kida@gmail.com} \\
\ \\
Yuichiro Urata \\
NTT Network Technology Laboratories \\
NTT Corporation \\
3-9-11 Midori-cho, Musashino-shi \\
Tokyo 180-8585 \\
Japan \\
\href{mailto:urata.yuichiro@lab.ntt.co.jp}{\tt urata.yuichiro@lab.ntt.co.jp} \\
\end{center}

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\begin{abstract}
We study a family of involutions on the space of sequences.
Many arithmetically or combinatorially interesting sequences appear
as eigensequences of the involutions.
We develop new tools for studying sequences using these involutions. 
\end{abstract}

\section{Introduction} \label{sec:1}
In his paper \cite{MR1362995}, Kaneko proved an interesting identity 
\begin{equation} \label{eq:1.1}
 \sum_{i=0}^{n+1} \binom{n+1}{i} (n+i+1) B_{n+i} =0
\end{equation}
for Bernoulli numbers $B_n$ defined by the exponential generating function
\begin{equation} \label{eq:1.2}
 \frac{x}{e^x-1} = \sum_{n=0}^{\infty} B_n \frac{x^n}{n!}.
\end{equation}
By using Kaneko's identity, we need about half the number of terms for 
computing $B_n$ compared with the usual identity
\[
 \sum_{i=0}^n \binom{n+1}{i} (-1)^i B_i = n+1.
\]
Kaneko proved the identity by means of a continued fraction expansion.
His paper also contains a sketch of 
another proof due to Zagier who uses an involutive
linear action on sequences.
Zagier's involution was further studied by various authors (see, for example,
\cite{MR2022347,MR1851531,MR1995582}).

The aim of this paper is
to define an infinite family of involutions on the space of
sequences, which generalize Zagier's involution,
and to prove various properties of the involutions and their eigensequences.
The definition of the involutions looks like that of modular forms
and this resemblance enables us to show that
generating functions of eigensequences
enjoy  properties analogous to modular forms.
The importance of these involutions is not only to provide
a tool to study general sequences but also to elicit a symmetry in
 classical and important sequences appearing as eigensequences.
In fact, as an application, we show that these involutions yield
many interesting identities involving the Bernoulli numbers, 
the Fibonacci numbers and so on.

This paper is organized as follows.
In the next section, we define the involutions and show their basic properties.
In Section \ref{sec:3}, we give examples of eigensequences 
of the involutions. It will become apparent that many arithmetically and/or 
combinatorially interesting sequences 
(namely, many core sequences in OEIS \cite{OEIS})
are eigensequences.
In Section \ref{sec:4}, we construct differential operators on the generating 
functions of eigensequences. 
In particular, we prove the existence of analogues of the Cohen-Rankin 
brackets.
These operators produce new eigensequences
from given eigensequences.
In Section \ref{sec:5.5}, we consider the action of the involutions 
on the endomorphism ring of the sequences.
By studying this action, we can produce infinitely many linear identities 
for eigensequences in Section \ref{sec:5}
including a generalization of Kaneko's identity. 

Throughout this paper, the binomial coefficient is a generalized one 
defined by
\[
 \binom{n}{k} = \frac{n (n-1) \cdots (n-k+1)}{k!}  
= \frac{n^{\underline{k}}}{k!}
\]
for $k \ge 0 $, where $n^{\underline{k}}$ is the falling factorial power 
defined by
\[
 n^{\underline{k}} = n (n-1) \cdots (n-k+1).
\]

\section{Definition and basic properties of involutions} \label{sec:2}
Let $\mathscr{S}$ be the set of sequences 
in a field $F$ of characteristic $0$.
Though we are primarily interested in integer or rational sequences,
we do not need such restriction to develop a general theory.
Thus we start with a general field $F$.

Since we have to deal with many sequences and their generating functions,
we need a more systematic notation than usual.
We denote the geometric series $\{ r^n \}$ of the ratio $r$ 
by $\langle r \rangle$ and also by $\boldsymbol{\delta}_n$ 
the sequence whose terms are all $0$ except for the $n$-th term which is $1$.
For any sequence $\boldsymbol{a} = \{ a_n \} \in \mathscr{S}$, 
we associate formal power series
\[
 G_i (\boldsymbol{a}, x) = \sum_{n=0}^{\infty} a_n \frac{x^n}{(n!)^i} \in 
F [[x]]  \quad (i=0,1).
\]
The series $G_0 (\boldsymbol{a},x)$ is the ordinary generating function and 
$G_1 (\boldsymbol{a}, x)$ is 
the exponential generating function of $\boldsymbol{a}$.
Further, we use the notation
\[
 [n]\boldsymbol{a} =a_n \text{ and } 
\left[ \frac{x^n}{(n!)^i}\right]  G_i (\boldsymbol{a}, x) = a_n.
\]
By the correspondence $\boldsymbol{a } \mapsto G_i (\boldsymbol{a} ,x )$,
we have isomorphisms of $F$-vector spaces:
\[
\varphi_i \ : \  \mathscr{S} \longrightarrow F[[x]].
\]
By these isomorphisms, we can endow $\mathscr{S}$ with 
 $F$-algebra structures.
Namely we can define products on $\mathscr{S}$ by
\[
[ n ]
 (\boldsymbol{a} \underset{i}{\ast} \boldsymbol{b})
= \sum_{j=0}^n \binom{n}{j}^i a_j b_{n-j}.
\]
The operation $\underset{0}{\ast}$ is called
the ordinary convolution and $\underset{1}{\ast}$ is the binomial convolution.
We set $\mathscr{S}_i =\varphi_i (\mathscr{S})$.

In addition, we denote the term-wise product sequence simply by 
$\boldsymbol{a}\boldsymbol{b}$, that is, 
$[n](\boldsymbol{a}\boldsymbol{b})=[n]\boldsymbol{a}\cdot [n]\boldsymbol{b}$.

\begin{definition} \label{def:2.1}
We define an action of a lower triangular $2 \times 2$ 
regular matrix $\displaystyle A= \begin{bmatrix}
							a &amp; 0 \\
							c &amp; d
						       \end{bmatrix}$ 
over $F$ on $\mathscr{S}_0$ and $\mathscr{S}$ by
\[
 G_0|_{[A]_k} (\boldsymbol{a}, x) = G_0 (\boldsymbol{a}|_{[A]_k}, x) 
= (c x +d)^{-k} G_0 \left( \boldsymbol{a}, \frac{ax}{cx+d} \right),
\]
where $k $ is an integer called the \textit{weight} of the action.
\end{definition}

It is plain to see that this operation satisfies
\begin{equation} \label{eq:2.1}
( G_0|_{[A]_k})|_{[B]_k} = G_0|_{[A B]_k} 
\end{equation}
for any such matrices $A$ and $B$.
Thus this is a well-defined action on $\mathscr{S}$.
The $n$-th term of the new sequence is explicitly given by
\begin{equation} \label{eq:2.2}
[n] (\boldsymbol{a}|_{[A]_k})  = \frac{1}{d^{n+k}} 
\sum_{j=0}^{n} \binom{n+k-1}{j} \ a_{n-j} \  a^{n-j} \ (-c)^{j}.
\end{equation}
If the order of $A$ in $\mathrm{GL_2 (F)}$ is $2$, then
we have an involution on $\mathscr{S}$.
Therefore we are particularly interested in the actions of 
the following matrices of order $2$:
\[
-I = \begin{bmatrix}
 -1 &amp; 0 \\
 0 &amp; -1 
\end{bmatrix}, \ 
  A_c = \begin{bmatrix}
       - 1 &amp; 0 \\
       c &amp;  1 
      \end{bmatrix} ,
\text{ and }
  -A_c = \begin{bmatrix}
        1 &amp; 0 \\
       -c &amp;  -1 
      \end{bmatrix} ,
\]
where $c $ is any element in $F$.
(Later at some places we take $c$ from a ring containing $F$).
For these matrices, the actions are given by
\begin{align}
[n] (\boldsymbol{a}|_{[-I]_k}) &amp; = (-1)^k a_n , \nonumber \\
[n] (\boldsymbol{a}|_{[A_c]_k}) &amp; =(-1)^n 
 \sum_{j=0}^n \binom{n+k-1}{j} \ a_{n-j} \ c^{j},  \label{eq:2.3} \\
[n] (\boldsymbol{a}|_{[-A_c]_k}) &amp; = (-1)^{n+k} 
     \sum_{j=0}^n \binom{n+k-1}{j} \ a_{n-j} \ c^{j} . \label{eq:2.4}
\end{align}
In particular, a special case
\[
[n] (\boldsymbol{a}|_{[A_{-1}]_1})  = 
 \sum_{j=0}^n \binom{n}{j} \ (-1)^j a_{j}  \\
\]
is the action which Zagier \cite{MR1362995} used.
Since the action of $-I$ is more or less trivial, we will not consider
it further and we assume that $A$ is one of $\pm A_{c}$ unless otherwise 
specified.
We write down the first few terms of $\boldsymbol{a}|_{[A_c]_k}$ for
later convenience:
\begin{equation} \label{eq:2.4.5}
 \begin{split}
 [0](\boldsymbol{a}|_{[A_c]_k}) &amp;=a_0;  \\
 [1](\boldsymbol{a}|_{[A_c]_k}) &amp;=  -k a_0 c-a_1;\\
 [2](\boldsymbol{a}|_{[A_c]_k}) &amp;= \binom{1+k}{2} \ a_0 c^2+(1+k) a_1
 c+a_2 ; \\
  [3](\boldsymbol{a}|_{[A_c]_k}) &amp;=  
 -\binom{2+k}{3}a_0 c^3-\binom{2+k}{2} a_1 c^2  -(2+k) a_2 c -a_3. 
 \end{split}
\end{equation}

The action of $[A]_k$ on $\mathscr{S}$ induces an action on $\mathscr{S}_1$.
\begin{proposition} \label{prop:2.2}
Let $\boldsymbol{a} \in \mathscr{S} $. For positive integers $k$, we have
\[
 G_1 (\boldsymbol{a}|_{[A]_k},x ) 
= \frac{1}{d^k} \frac{d^{k-1}}{dx^{k-1}} 
\left( \exp \left( -\frac{c}{d} x\right)
 \underbrace{\idotsint}_{k-1}  G_1 \left(\left\langle 
\frac{a}{d}\right\rangle \boldsymbol{a} , 
 x \right) \underbrace{dx \cdots dx}_{k-1} \right),
\]
where all the integral sign means the formal integration from $0$ to $x$.
\end{proposition}
\begin{proof}
We compute
\begin{align*}
 G_1 (\boldsymbol{a}|_{[A]_k},x )  &amp;= \frac{1}{d^k} 
\sum_{n \ge 0} \frac{1}{d^n} \left( \sum_{j=0}^n  \binom{n+k-1}{j}
 (-c)^j a^{n-j} a_{n-j}\right) \frac{x^n}{n!} \\
&amp;= \frac{1}{d^k}  \sum_{n \ge 0} \frac{(n+k-1)!}{n!}
\left( \sum_{j=0}^n \binom{n}{j} \left( -\frac{c}{d}\right)^j 
\left( \frac{a}{d}\right)^{n-j} a_{n-j} (n-j)! \frac{1}{(n-j+k-1)!}\right)
\frac{x^n}{n!}.
\end{align*}
The inner sum is the binomial convolution of
\[
 \left\langle -\dfrac{c}{d} \right\rangle \quad \text{and} \quad 
\left(\dfrac{a}{d}\right)^n a_n \dfrac{n!}{(n+k-1)!}.
\]
The exponential generating function of the latter sequence is 
\[
x^{1-k} \underbrace{\idotsint}_{k-1}  G_1  
\left(\left\langle 
\frac{a}{d}\right\rangle \boldsymbol{a} , 
 x \right) \underbrace{dx \cdots dx}_{k-1}.
\]
This completes the proof of the proposition.
 \end{proof}
The special case where $k=1$ will be important for us:
\begin{equation} \label{eq:2.5}
 G_1 (\boldsymbol{a}|_{[A]_1},x ) = \frac{1}{d} 
\exp \left( -\frac{c}{d} x \right)
G_1 \left( \left\langle 
\frac{a}{d}\right\rangle \boldsymbol{a} , x \right) .
\end{equation}

\begin{remark} \label{rmk:2.3}
 We can prove the following formula similarly for $k \le 0$:
\[
\left[ \frac{x^n}{n!}\right]  G_1 (\boldsymbol{a}|_{[A]_k},x ) 
= \left[ \frac{x^n}{n!}\right]  \frac{1}{d^k} 
 \underbrace{\idotsint}_{1-k} \left( 
\exp \left( -\frac{c}{d} x \right)
 \frac{d^{1-k}}{dx^{1-k}} G_1 
\left(\left\langle \frac{a}{d} \right\rangle \boldsymbol{a}  ,x \right) \right)
\underbrace{dx \cdots dx}_{1-k} 
\]
provided $n \ge 1-k $.
\end{remark}

Now we define the main object of our study.
\begin{definition} \label{def:2.4}
Let $A$ be a lower triangular regular matrix of order $2$.
A sequence $\boldsymbol{a}$ satisfying 
\[
 \boldsymbol{a}|_{[A]_k}=  s \boldsymbol{a}
\]
with some $s \in \{\ \pm 1\}$ is called an \textit{eigensequence} of $[A]_k$.
We call $s$ the \textit{sign} of the eigensequence. 
We denote the eigenspaces by
\[
 \mathscr{S}(A)_k^{+} 
 = \left\{ \boldsymbol{a} \in \mathscr{S} \ : \ \boldsymbol{a}|_{[A]_k}
= + \boldsymbol{a} \right\} 
\text{ and } 
 \mathscr{S}(A)_k^{-} 
 = \left\{ \boldsymbol{a} \in \mathscr{S} \ : \ \boldsymbol{a}|_{[A]_k}
= - \boldsymbol{a} \right\}. 
\]
\end{definition}

Although a resemblance between this definition and that of modular forms 
is clear (see \cite{MR2289048} for example), several differences 
will appear in the following. Here we only note that each space
$\mathscr{S} (A)_k^{\pm}$ is an infinite-dimensional $F$-vector space
for every integer $k$.

By \eqref{eq:2.2} we have
\begin{equation} \label{eq:2.6}
 \boldsymbol{a} \in \mathscr{S}(A)_k^{\pm}
\Leftrightarrow \left(\pm 1 - \frac{a^n}{d^{n+k}}  \right) a_n
= \frac{1}{d^{n+k}}\sum_{j=1}^{n} \binom{n+k-1}{j}a_{n-j} a^{n-j} (-c)^j.
\end{equation}
If $A= \pm A_c$, then
$\left(\pm 1 - \frac{a^n}{d^{n+k}}  \right)$ equals $\pm 2 $ or $0$ 
alternatively with $n$. This means that we can choose one of the 
two consecutive terms freely and 
the next term is determined automatically by the preceding 
terms.
We also define
\[
 \mathscr{S}_i (A)_k^{\pm} = \varphi_i ( \mathscr{S}(A)_k^{\pm} ).
\]

When $A=A_0$, the eigensequences are easy to describe.

\begin{proposition} \label{prop:2.5} 
The following equivalences hold for $i=0,1$:
\begin{align*}
 \boldsymbol{a} \in \mathscr{S}(A_0)_k^+ &amp; \Longleftrightarrow 
               G_i (\boldsymbol{a},x ) \text{ is an even power series}; \\
 \boldsymbol{a} \in \mathscr{S}(A_0)_k^- &amp; \Longleftrightarrow 
               G_i (\boldsymbol{a},x ) \text{ is an odd power series}. 
\end{align*}
\end{proposition}
\begin{proof}
 If $A=A_0$, then 
Definition \ref{def:2.1} implies that
$\boldsymbol{a} \in \mathscr{S} (A_0)_k^{\pm}$ if and only if
\[
 G_0 (\boldsymbol{a},-x) = \pm G_0 (\boldsymbol{a},x).
\]
This means that $ G_0 (\boldsymbol{a},x)$ is an odd or even 
power series according to the sign of $\boldsymbol{a}$.
Moreover, $ G_0 (\boldsymbol{a},x)$ is an odd (resp. even) 
power series if and only if $ G_1 (\boldsymbol{a},x)$ is an odd (resp. even)
power series.
This completes the proof.
\end{proof}

\begin{example} \label{ex:2.5.1}
 The sequence of the Euler numbers $\boldsymbol{E}=\{E_n \}$ 
(see \cite[p.559]{MR1397498}) is an example 
in $\mathscr{S} (A_0)_k^{+}$, because it is defined by
\begin{equation} \label{eq:2.7.5}
 G_1 (\boldsymbol{E}, x) = \sec x = 
\sum_{n=0}^{\infty} E_{2n} \frac{x^{2n}}{(2n)!}.
\end{equation}
An example from $\mathscr{S} (A_0)_k^{-}$ is the sequence of the tangent numbers $\boldsymbol{T}=\{ T_n \}$ defined by
\[
 G_1 (\boldsymbol{T}, x) = \tan x = 
\sum_{n=0}^{\infty} T_{2n+1} \frac{x^{2n+1}}{(2n+1)!},
\]
which have a close relation to the Bernoulli numbers \cite[p.287]{MR1397498}.
\end{example}

Also we can deduce the following identities readily from \eqref{eq:2.3}
and \eqref{eq:2.4}.

\begin{proposition} \label{prop:2.6} 
We have
 \[
  \mathscr{S}(A_c)^{\pm}_k = \begin{cases}
	      \mathscr{S} (-A_{c})_k^{\pm}, &amp; \text{ if $k$ is even}; \\
	      \mathscr{S} (-A_{c})_k^{\mp}, &amp; \text{ if $k$ is odd}.
			     \end{cases}
 \]
\end{proposition}

By these propositions, we shall mainly consider the eigenspaces
$\mathscr{S}(A_c)^{\pm}_k$
with $c \neq 0$.

As in the case of modular forms, the space of eigensequences has a 
graded structure.

\begin{proposition} \label{prop:2.7}
 If $\boldsymbol{a} \in  \mathscr{S}(A)_{k_1}^{+} $
 and $\boldsymbol{b} \in  \mathscr{S}(A)_{k_2}^{+} $, then
$\boldsymbol{a} \underset{0}{\ast} 
\boldsymbol{b} \in \mathscr{S}(A)_{k_1+k_2}^{+} $.
If $A \neq \pm A_0$, then
$\displaystyle \mathscr{S}(A)^{+} = \oplus_{k \in \Z }
 \mathscr{S}(A)_{k}^{+}$
is a graded ring under this product.
\end{proposition}
\begin{proof}
Let $\boldsymbol{a} \in  \mathscr{S}(A)_{k_1}^{+} $
 and $\boldsymbol{b} \in  \mathscr{S}(A)_{k_2}^{+} $. Then we 
have
\begin{align*}
 G_0|_{[A]_{k_1 + k_2}} 
( \boldsymbol{a} \underset{0}{\ast} \boldsymbol{b} ,x ) 
&amp; = (G_0 (\boldsymbol{a},x) G_0 (\boldsymbol{b} , x))|_{[A]_{k_1+k_2}} \\
&amp;= (cx+d)^{-(k_1 +k_2)} G_0 \left( \boldsymbol{a}, \frac{ax}{cx+d} \right)
   G_0 \left( \boldsymbol{b}, \frac{ax}{cx+d} \right) \\
&amp;= G_0 ( \boldsymbol{a}|_{[A]_{k_1}}, x ) G_0 ( \boldsymbol{b}|_{[A]_{k_2}}, x ) \\
&amp;= G_0 ( \boldsymbol{a}, x ) G_0 ( \boldsymbol{b}, x ) \\
&amp;=  G_0 ( \boldsymbol{a} \underset{0}{\ast} \boldsymbol{b} ,x ).
\end{align*}
This shows the first assertion.
To show the second assertion, it suffices to prove that if $k_1 \neq k_2$,
then $\mathscr{S}(A)_{k_1}^{+} \cap \mathscr{S}(A)_{k_2}^{+} = \{ 0 \}$.
Let $\boldsymbol{a} \in \mathscr{S}(A)_{k_1}^{+} \cap \mathscr{S}(A)_{k_2}^{+}$.
Then 
\[
 (cx+d)^{k_1} G_0 \left( \boldsymbol{a}, \frac{ax}{cx+d} \right)
= (cx +d )^{k_2} G_0 \left( \boldsymbol{a}, \frac{ax}{cx+d} \right)
\]
holds.
Since $F[[x]]$ is an integral domain, we have $k_1=k_2$ provided $c \neq 0$.
\end{proof}


In a similar manner, we can show the following proposition.
\begin{proposition} \label{prop:2.9}
Let $s_1, s_2 \in \{ \pm 1 \}$.
If $\boldsymbol{a} \in  \mathscr{S}(A)_{k_1}^{s_1} $
 and $\boldsymbol{b} \in  \mathscr{S}(A)_{k_2}^{s_2} $, then
$\boldsymbol{a} \underset{0}{\ast} \boldsymbol{b} 
\in \mathscr{S}(A)_{k_1+k_2}^{s_1 s_2} $.
\end{proposition}

By these propositions, it is clear that considering these involutions
with varying $k$ rather than a fixed $k$.

A special case of Proposition \ref{prop:2.9} implies the following.
\begin{proposition} \label{prop:2.9.5}
There exists an isomorphism 
\[
 \mathscr{S}(A_c)_k^{+} \cong  \mathscr{S}(A_c)_{k-1}^{-},
\quad \boldsymbol{a} \mapsto \boldsymbol{\delta}_1 \underset{0}{\ast} \boldsymbol{a}
\]
of vector spaces for every integer $k$.
\end{proposition}
\begin{proof}
Since $G_0 (\boldsymbol{\delta}_1,x)=x$, we can see 
$\boldsymbol{\delta}_1 \in \mathscr{S}(A_c)_{-1}^{-}$. It follows from
 Proposition \ref{prop:2.9} that the
 convolution $\boldsymbol{\delta}_1 \underset{0}{\ast} \boldsymbol{a}$
belongs to $\mathscr{S}(A_c)_{k-1}^{-}$.
The given map is obviously an injective linear map. To show the surjectivity,
 we note that the sequence $\boldsymbol{\delta}_1 \underset{0}{\ast} \boldsymbol{a}$
 is a shift of $\boldsymbol{a}$ to the right.
Therefore it is enough to prove that every sequence in
 $\mathscr{S}(A_c)_{k-1}^{-}$ begins with $0$.
This assertion follows immediately from \eqref{eq:2.4.5}.
\end{proof}

In terms of
generating functions, Proposition \ref{prop:2.9.5} 
gives the following isomorphism:
\[
  \mathscr{S}_0(A)_k^{+} \cong  \mathscr{S}_0(A)_{k-1}^{-},
\quad G_0 (\boldsymbol{a},x) \mapsto x G_0 (\boldsymbol{a},x).
\]
It is worth while to remark that
the same correspondence gives an injective linear map
from $\mathscr{S}(A_c)_k^{-} $ to $\mathscr{S}(A_c)_{k-1}^{+}$ but this
is not surjective in general.


There is a variant of Proposition \ref{prop:2.9} for binomial convolution.
Before we state it, we note the following equivalence.
If $k=1$, then by \eqref{eq:2.5} we see
\begin{equation} \label{eq:2.8}
 \boldsymbol{a} \in \mathscr{S}(A_c)_1^{\pm}
\Longleftrightarrow
G_1 (\boldsymbol{a}, x) 
= \pm \exp (-cx) G_1 (\langle -1 \rangle \boldsymbol{a},x ),
\end{equation}
which is a generalization of \cite[Theorem 3.2]{MR1851531}.
Observe here that we have an identity
\[
 G_1 (\langle -1 \rangle \boldsymbol{a},x ) = 
G_1 (\boldsymbol{a}, -x ).
\]

\begin{proposition} \label{prop:2.10}
Let $s_1,s_2 \in \{\pm 1\}$.
  If $ \boldsymbol{a} \in \mathscr{S} (A_{c_1})_1^{s_1}$ and 
$\boldsymbol{b} \in \mathscr{S} (A_{c_2})_1^{s_2}$, then
\[
\boldsymbol{a} \underset{1}{\ast} \boldsymbol{b} 
 \in \mathscr{S} (A_{c_1+ c_2})_1^{s_1 s_2}.
\]
\end{proposition}
\begin{proof}
By \eqref{eq:2.8}, we have
\begin{align*}
G_1 (\boldsymbol{a}, x) 
&amp; = s_1 \exp (-c_1 x) G_1 (\langle -1 \rangle \boldsymbol{a}, x ), \\
G_1 (\boldsymbol{b}, x) 
&amp; = s_2 \exp (-c_2 x) G_1 (\langle -1 \rangle \boldsymbol{b}, x ).
\end{align*}
Multiplying the both sides,
we obtain
\[
G_1 (\boldsymbol{a}, x) G_1 (\boldsymbol{b}, x) 
 = s_1 s_2 \exp (-(c_1+c_2)x) 
G_1 (\langle -1 \rangle \boldsymbol{a}, x )
G_1 (\langle -1 \rangle \boldsymbol{a}, x ).
\]
The left hand side is equal to $G_1 (\boldsymbol{a} \underset{1}{\ast}
\boldsymbol{b}, x  ) $. From the right hand side, we get
\[
G_1 (\langle -1 \rangle \boldsymbol{a}, x )
G_1 (\langle -1 \rangle \boldsymbol{a}, x )
 = G_1 (\boldsymbol{a}, -x) G_1 (\boldsymbol{b}, -x) 
 = G_1 (\boldsymbol{a} \underset{1}{\ast} \boldsymbol{b}, -x  ) 
 = G_1 (\langle -1 \rangle (\boldsymbol{a} 
\underset{1}{\ast} \boldsymbol{b}), x  ) .
\]
Now the proposition follows from \eqref{eq:2.8}.
\end{proof}

Both Propositions \ref{prop:2.9} and \ref{prop:2.10} will be 
generalized in a broader context in Section \ref{sec:4}
(see Theorems \ref{thm:4.1} and \ref{thm:4.8}).

If the $0$-th term of a sequence $\boldsymbol{a} \in \mathscr{S}$ 
is not $0$, 
then $G_i (\boldsymbol{a},x ) \ (i=0,1)$ are invertible in $F[[x]]$.

\begin{corollary} \label{cor:2.8}
 If $\boldsymbol{a} \in \mathscr{S}(A_c)_k^{+}$ and $[0]\boldsymbol{a}
 \neq 0$, then we have
\[
 G_0 (\boldsymbol{a},x )^{-1} \in \mathscr{S}_0 (A_c)_{-k}^{+}.
\]
Moreover, if $k=1$, then we have
\[
 G_1 (\boldsymbol{a},x )^{-1} \in \mathscr{S}_1 (A_{-c})_{1}^{+}. 
\]
\end{corollary}
\begin{proof}
If the generating functions are invertible, then
$G_i (\boldsymbol{a}, x) G_i (\boldsymbol{a},x)^{-1} =1 =
G_i (\boldsymbol{\delta}_0,x)$.
Since $\boldsymbol{\delta}_{0}$ belongs to $\mathscr{S}
 (A_c)_0^{+} \cap \mathscr{S} (A_0)_k^{+}$ for all $c$ and $k$,
the corollary follows from Propositions \ref{prop:2.9} and \ref{prop:2.10}.
\end{proof}

The following propositions give easy ways to alter $c$.

\begin{proposition} \label{prop:2.11} 
If $\boldsymbol{a} \in \mathscr{S} (A_c)_k^{\pm}$, then
$\langle r \rangle \boldsymbol{a} \in \mathscr{S} (A_{cr})_k^{\pm}$.
\end{proposition}
We omit the proof since it is straightforward.

\begin{proposition} \label{prop:2.12} 
 If $\boldsymbol{a} \in \mathscr{S} (A_{c_1})_k^{\pm}$,
then $ \boldsymbol{a}|_{[A_{c_2}]_k} \in \mathscr{S}
 (A_{2c_2-c_1})_k^{\pm}$.
\end{proposition}
\begin{proof}
 Using \eqref{eq:2.1} and the relation
$A_{c_2} A_{2c_2-c_1}=A_{c_1} A_{c_2}$, we compute
\[
 (\boldsymbol{a}|_{[A_{c_2}]_k} )_{[A_{2c_2-c_1}]_k} =
 \boldsymbol{a}|_{[A_{c_2} A_{2c_2-c_1} ]_k}
= \boldsymbol{a}|_{[A_{c_1} A_{c_2} ]_k}
=( \boldsymbol{a}|_{[A_{c_1}]_k} )_{[A_{c_2}]_k}
=\pm \boldsymbol{a}|_{[A_{c_2}]_k}.
\]
This proves the assertion.
\end{proof}

We close this section with the following remark.
For each $A$ of order $2$ and each $k$, we have projections
\begin{align*}
 \pi (A)_k^{+} &amp;= \frac{1 + [A]_k }{2} \ : \ \mathscr{S} 
\longrightarrow  \mathscr{S}(A)_k^{+}  \\
\intertext{and}
 \pi (A)_k^{-} &amp;= \frac{1 - [A]_k }{2} \ : \ \mathscr{S} 
\longrightarrow  \mathscr{S}(A)_k^{-}.  
\end{align*}
Also by Proposition \ref{prop:2.7}, 
if $A \neq \pm A_0$, then there is a surjection 
\[
 \pi (A) = \bigoplus_{k \in \Z} \pi (A)_k^+ \ : \ \mathscr{S} \longrightarrow 
\mathscr{S}(A)^{+} = \bigoplus_{k \in \Z} \mathscr{S}(A)_{k}^{+}.
\]
The kernel of $\pi (A)$ is $\cap_{k} \mathscr{S}(A)_{k}^{-}$.
In a similar manner as in the proof of Proposition \ref{prop:2.7},
it is shown that it is actually $0$. 
Hence the surjection is an isomorphism.

\section{Eigensequences} \label{sec:3}
In this section, we give various examples of 
sequences in $\mathscr{S}(A_c)_k^{\pm}$ (see Propositions \ref{prop:2.5}
and \ref{prop:2.6}).
Although it is easy to construct such sequences from arbitrary sequences 
using the projections $\pi (A)_k^{\pm}$, 
almost all sequences thus obtained are unnatural and useless.
Therefore we systematically search interesting eigensequences 
in the following.
In the course of searching, 
we can reveal the structure of $\mathscr{S}(A_c)_k^{\pm}$ 
(Theorem \ref{thm:3.3.5}).
Note that Z.-H. Sun \cite[Section 2]{MR1851531} found some examples 
in $\mathscr{S}(A_{-1})_{1}^{+}$ and some of
the following examples are generalizations of his.

We begin with necessary conditions for a given sequence 
to be an eigensequence.

\begin{lemma} \label{lem:3.1}
Let $\boldsymbol{a} \in \mathscr{S}$.
\begin{enumerate}
 \item \label{lem:3.1.1} If $\boldsymbol{a} \in \mathscr{S}(A_c)_k^+ $, then
\begin{align}
 a_1 &amp;= -\frac{1}{2} kc a_0,  \label{eq:3.1} \\
 a_3 &amp; = \frac{1}{24}c(2+k)(-12 a_2 + a_0 c^2 k +a_0 c^2 k^2). \nonumber
\end{align}
Moreover $k$ is an integer solution of the quadratic equation
\begin{equation} \label{eq:3.2}
 (a_1^3 +3 a_0^2 a_3 -3 a_0 a_1 a_2 ) k^2 + (3a_1^3 - 6 a_0 a_1 a_2)k
 +2 a_1^3 =0.
\end{equation}
 \item \label{lem:3.1.2} If $\boldsymbol{a} \in \mathscr{S}(A_c)_k^- $, then
\begin{align}
 a_0 &amp;= 0, \label{eq:3.3} \\
 a_2 &amp;= -\frac{1}{2} (1+k) c a_1 , \label{eq:3.4} \\
 a_4 &amp; = \frac{1}{24} c(3+k) (-12 a_3 + a_1 c^2 (k+1) + a_1  
c^2 (k+1)^2). \label{eq:3.5}
\end{align}
Moreover $k$ is an integer solution of the quadratic equation
\[ (a_2^3 +3 a_1^2 a_4 - 3 a_1 a_2 a_3 )k^2 + 
(6 a_1^2 a_4  -12 a_1 a_2 a_3 +5 a_2^3) k 
+3 a_1^2 a_4 +6 a_2^3 - 9 a_1 a_2 a_3 =0. \]
\end{enumerate}
\end{lemma}
\begin{proof}
The conditions on each term is obtained directly from the 
identities following \eqref{eq:2.4.5} 
(see also \eqref{eq:2.6}).
By eliminating $c$ from these equalities, we obtain the equations for $k$.

Note that the latter half \ref{lem:3.1.2} also follows from the isomorphism
\[
 \mathscr{S_c}(A_c)_{k+1}^{+} \cong  \mathscr{S_c}(A_c)_{k}^{-}
\]
in Proposition \ref{prop:2.9.5}.
In other words, the relations \eqref{eq:3.3}, \eqref{eq:3.4} and
\eqref{eq:3.5} can be obtained by a right shift of the relations 
in \ref{lem:3.1.1} and a substitution of $k$ by $k+1$.
\end{proof}

Using this lemma, we can decide that a given sequence is not an
eigensequence or 
determine candidates for  $c$ and $k$.

We first give some examples of eigensequences 
with polynomial generating functions, namely finite eigensequences.

\begin{example}[Polynomial generating functions] \label{ex:3.2}
If the generating function $G_0 (\boldsymbol{a},x )$ for
 $\boldsymbol{a} \in \mathscr{S} (A_c)_k^{\pm} \ (c \neq 0)$ is 
a polynomial of degree $m \ge 1$,
then it is easy to see that the weight $k$ should be negative and
equal or less than $-m$.
If $k$ is exactly equal to $-m$, then we can show that $[A_c]_{-m}$ defines an
involution on the subspace
\[
 F[x]_m = \{ G(x) \in F[x] \ : \ \deg G(x) \le m \}.
\]
 By Lemma \ref{lem:3.1}, we have an inequality
\[
 \dim_F (\mathscr{S}_0 (A_c)_k^{+} \cap F[x]_m )
\ge \dim_F (\mathscr{S}_0 (A_c)_k^{-} \cap F[x]_m ).
\]
This implies 
\[
  \dim_F (\mathscr{S}_0 (A_c)_k^{+} \cap F[x]_m ) = 
\left\lceil \dfrac{m+1}{2} \right\rceil, \quad 
  \dim_F (\mathscr{S}_0 (A_c)_k^{-} \cap F[x]_m ) = 
\left\lfloor \dfrac{m+1}{2} \right\rfloor.
	     \]
As $k$ gets smaller, we have additional constraints on the terms
of eigensequences. Therefore the dimension gets smaller, too.

Suppose that we want to find generating functions of degree $1$ for
plus eigensequences. Setting $a_2 =a_3 = 0 $ in \eqref{eq:3.2}, we have
\[
 a_1^3 k^2 +3a_1^3 k +2 a_1^3 =0.
\]
This equation yields that $k=-1$ or $-2$. 
From \eqref{eq:3.1} it follows that $a_1=c a_0/2$ or
$c a_0$. We can easily show that these are indeed eigensequences.
Similar calculation leads to the following polynomial generating functions:
\begin{itemize}
\setlength\itemindent{35pt}
 \item[$\deg G_0 =1$] 
\begin{align*}
   1+ \frac{c}{2}x &amp; \in \mathscr{S}_0(A_c)_{-1}^+ ,
        &amp; x  &amp; \in \mathscr{S}_0(A_c)_{-1}^{-} \cap \mathscr{S}_0(A_0)_{k}^{-}, \\
       1+cx &amp; \in \mathscr{S}_0(A_c)_{-2}^+ ;
                &amp;  &amp;   
\end{align*}
 \item[$\deg G_0 =2$]
\begin{align*}
   t +ct x + sx^2 &amp; \in \mathscr{S}_0(A_c)_{-2}^+ ,
             &amp; x \left( 1+ \frac{c}{2}x \right) 
                        &amp; \in \mathscr{S}_0(A_c)_{-2}^{-} , \\ 
  (1+cx)\left( 1+\frac{c}{2}x \right) &amp; \in \mathscr{S}_0(A_c)_{-3}^{+} ,
            &amp; x(1+cx) &amp; \in \mathscr{S}_0(A_{c})_{-3}^{-} \\
  (1+cx)^2 &amp; \in \mathscr{S}_0(A_c)_{-4}^{+} ;
    &amp;  &amp; 
 \end{align*}
\end{itemize}
where $s \text{ and } t$ are arbitrary parameters.

Since 
$\boldsymbol{\delta}_{1}  \in \mathscr{S}(A_c)_{-1}^{-}$, 
we can prove that
\[
G_0( \boldsymbol{\delta}_{m},x) =  x^m =
G_0 ( \boldsymbol{\delta}_{1} ,x )^m \in \mathscr{S}_0 (A_c)_{-m}^{(-1)^m}
\]
for positive integer $m$ by using Proposition \ref{prop:2.9} repeatedly.
Combining this with Proposition \ref{prop:2.5}, we conclude 
\begin{equation} \label{eq:3.0}
\boldsymbol{\delta}_m \in \mathscr{S} (A_c)_{-m}^{(-1)^m}
 \cap
\mathscr{S} (A_0)_{k}^{(-1)^m} 
\end{equation}
for all $c$ and $k$.
\end{example}

These finite eigensequences themselves are not so interesting, but
they provide primary ingredients of general eigensequences.
We shall see this in the following rational generating functions.

\begin{example}[Rational generating functions] \label{ex:3.3}
Combining polynomial generating functions in various ways,
 we have rational generating functions of eigensequences, 
whose weights can be computed
 using Proposition \ref{prop:2.9} and Corollary \ref{cor:2.8}.
The simplest examples are
\[
 G_0 (\langle c \rangle ,x ) 
= \frac{1}{1-cx} \in \mathscr{S}_0 (A_{-2c})^+_{1} \cap
 \mathscr{S}_0 (A_{-c})^+_{2}. 
\]

We also have
\[
 G_0 (\boldsymbol{a}, x) 
   = \frac{2+cx}{t+tcx+sx^2} \in \mathscr{S}_0 (A_c)_1^+ , \quad
 G_0 (\boldsymbol{b}, x) 
  = \frac{x}{t+tcx+sx^2} \in \mathscr{S}_0 (A_c)_1^- .
\]
As a special case of $\boldsymbol{a}$, we have
the Lucas sequence  $\boldsymbol{L}=\{ L_n \}$  defined by the recurrence
\[
 L_0 = 2, \ L_1 = 1, \ L_n = L_{n-1} + L_{n-2}.
\]
In fact, since $ G_0 (\boldsymbol{L},x)= \displaystyle \frac{2-x}{1-x-x^2} $, 
we have $ \boldsymbol{L} \in \mathscr{S} (A_{-1})_1^{+} $.
As for $\boldsymbol{b}$, we have 
 the Fibonacci sequence 
$\boldsymbol{F}=\{F_n \}$ defined by 
\[
 F_0=0, \ F_1 =1, \ F_n = F_{n-1}+F_{n-2},
\]
whose generating function is 
$ \displaystyle G_0 (\boldsymbol{F}, x) = \frac{x}{1-x-x^2} 
\in \mathscr{S}_0 (A_{-1})_1^{-} $.
This means that $\boldsymbol{F}$ and $\boldsymbol{L}$ are not 
only a basis of the 2-dimensional vector space of sequences 
satisfying the recurrence $a_{n+1}=a_{n}+a_{n-1}$, but also
a basis as a $\langle [A_{-1}]_1 \rangle$-module.
This fact plays an essential role 
in the researches of Fibonacci numbers although it is not noticed 
explicitly.

The binomial coefficients $\displaystyle  \left\{ \binom{n+k-1}{n}
 \right\}_{n \ge 0} $ 
are also of this type, since the ordinary generating function is 
$\displaystyle (1-x)^{-k} $. Therefore we see
\[
\left\{ \binom{n+k-1}{n}
 \right\}_{n \ge 0}
\in \mathscr{S} (A_{-1})^{-}_{2k}.
\]
\end{example}

The generating function
\[
 \mathscr{E}_k (x) = \frac{1}{\left(1+\dfrac{c}{2}x\right)^k}
                   = \sum_{n=0}^{\infty } \binom{n+k-1}{n} 
\left( -\frac{c}{2}\right)^n x^n \in \mathscr{S}_0 (A_c)_k^{+}
\]
plays a special role.

\begin{theorem} \label{thm:3.3.5}
We have a direct sum decomposition
\[
 \mathscr{S}_0 (A_c)_k^{+} = F\cdot  \mathscr{E}_k (x) \oplus
x^2  \mathscr{S}_0 (A_c)_{k+2}^{+}
\]
for any integer $k$.
Moreover, any generating function  $G_0 (\boldsymbol{a},x)
 \in  \mathscr{S}_0 (A_c)_k^{+}$
can be written uniquely as a sum of the form
\begin{equation} \label{eq:3.7}
G_0 ( \boldsymbol{a},x ) = \sum_{i=0}^{\infty} \alpha_{2i} x^{2i} 
 \mathscr{E}_{k+2i} (x) \quad (\alpha_{2i} \in F). 
\end{equation}
If $G_0 (\boldsymbol{a},x) \in  \mathscr{S}_0 (A_c)_k^{-}$, then
it can be written uniquely as 
\begin{equation} \label{eq:3.8}
 G_0 (\boldsymbol{a},x) = \sum_{i=0}^{\infty} \alpha_{2i+1} x^{2i+1} 
 \mathscr{E}_{k+2i+1} (x) \quad (\alpha_{2i+1} \in F). 
\end{equation}
\end{theorem}
\begin{proof}
 Consider the map $g: \mathscr{S}_0 (A_c)_k^{+} \longrightarrow F$ 
defined by $G_0 ( x) \mapsto G_0 (0) $. The dimension of the image of $g$ 
is at most $1$.
Since $[x^0]  \mathscr{E}_{k} (x) \neq 0$, the dimension is $1$, indeed.
Suppose that $G_0 (x) \in \ker g $.
Then $[x] G_0 (x) $ is also $0$ by \eqref{eq:3.1}.
We obtain a factorization $G_0 (x) =x^2 H (x)$.
From \eqref{eq:3.0}, we have $H(x) \in \mathscr{S}_0 (A_c)_{k+2}^{+}$.
This proves the first assertion.
Unless $H(x)$ is zero or a constant, we can 
continue the same argument for $H(x)$, we can expand $G_0 (x)$ 
like \eqref{eq:3.7}.
If $H(x)$ is zero or a constant, the process stops and obtain
a finite sum expansion.
The equation \eqref{eq:3.8} follows from \eqref{eq:3.7}
by using the isomorphism in Proposition \ref{prop:2.9.5}.
\end{proof}

An explicit formula for $\alpha_{i}$ in \eqref{eq:3.7} and \eqref{eq:3.8}
will be given later (Proposition \ref{prop:3.10.5}).

This theorem provides more precise test for eigensequences than Lemma 
\ref{lem:3.1}. 
But we should note that it is still difficult to extract interesting sequences 
from this structure theorem. That being so, we continue our quest.

\begin{example} \label{ex:3.4}
 In this example, we deal with irrational generating functions.
Let $G_0 (x) \in \mathscr{S}_0$ be an invertible power series.
If we choose a square root $\sqrt{a_0}$ of $a_0 = G_0 (0)$, then 
$\sqrt{G_0 (x)}$ is uniquely determined and belongs to
$\mathscr{S}_0 \otimes_F F(\sqrt{a_0})$.
Further if $G_0 (x) \in \mathscr{S}_0 (A_{c})^{+}_{2k} $, then it is
 easy to verify that
$\sqrt{G_0 (x)} \in \mathscr{S}_0 (A_{c})^{+}_{k} \otimes_F F(\sqrt{a_0})$.
As an example, since $\displaystyle \frac{1}{1-4x}  
\in \mathscr{S}_0 (A_{-4})_{2}^{+}$,
we have
\[
 \frac{1}{\sqrt{1-4x}} \in \mathscr{S}_0 (A_{-4})_1^{+},
\]  
whose $n$-th 
coefficient is $\displaystyle \binom{2n}{n}$.
Let $\boldsymbol{C}=\{C_n \}$ be 
the Catalan numbers (\cite[6.2.1 Proposition]{MR1676282}) 
whose generating function is known to be
\[
G_0 (\boldsymbol{C},x) = \frac{1-\sqrt{1-4x}}{2x} .
\]
We can show that $G_0 (\boldsymbol{C},x)^2 \in \mathscr{S}_0 (A_{-4})_1^{+}$.
Since it is well known that $G_0 (\boldsymbol{C} , x)^2 
= \dfrac{G_0 (\boldsymbol{C} , x)-1}{x} $, we have $\{ C_{n+1} \}_{n \ge 0} 
\in \mathscr{S}(A_{-4})_1^{+}$.

The generating function of Motzkin numbers is another example:
\[
 \frac{1-x-\sqrt{1-2x-3x^2}}{2x} \in \mathscr{S}_0 (A_{-2})_0^{-}.
\]
\end{example}

\begin{example} \label{ex:3.4.5}
There are also examples of transcendental generating functions.
Let $\boldsymbol{H}$ be the sequence of harmonic numbers 
(\cite[Section 6.3]{MR1397498}):
\[
[0]\boldsymbol{H} =0 , \quad  
[n] \boldsymbol{H} = 1 + \frac{1}{2} + \cdots + \frac{1}{n}.
\]
We have
\[
 G_0 (\boldsymbol{H},x ) = \frac{1}{1-x} \log \left( \frac{1}{1-x} \right)
\in \mathscr{S}_0(A_{-1})^{-}_{2}.
\]
More generally, Pfaff's reflection law for the Gaussian hypergeometric 
function (\cite[(5.101)]{MR1397498})
\[
 \frac{1}{(-x+1)^{\alpha}} F \left( \left.
\genfrac{}{}{0pt}{0}{\alpha ,\beta }{\gamma } 
\right|  \frac{-x}{-x+1}  \right)
= F \left( \left.
\genfrac{}{}{0pt}{0}{\alpha ,\gamma - \beta }{\gamma } 
\right| x   \right)
\]
implies 
\[
F \left( \left.
\genfrac{}{}{0pt}{0}{\alpha ,\beta }{2 \beta } 
\right| x   \right) \in \mathscr{S}_0 (A_{-1})_{\alpha}^{+}.
\]
\end{example}

 We next proceed to find examples of 
exponential generating functions of eigensequences
using \eqref{eq:2.8}.

\begin{example} \label{ex:3.5}
We begin with examples of polynomials of $\exp (x)$.
It is possible to determine all such polynomials giving eigensequences.
First of all, a monomial generating function $\exp (cx)$ gives
\[
 \langle c \rangle \in \mathscr{S} (A_{-2c})_1^+
\]
as we saw in Example \ref{ex:3.3}.

We say that a polynomial $P(t) \in F[t]$ is self-reciprocal if 
$u^{\deg P} P(t/u) \in F[t,u]$ is a symmetric polynomial:
namely, setting $d=\deg P$, it satisfies
\[
 u^d P(t/u) =t^d P(u/t).
\]
Also a polynomial $P(t)$ is called skew self-reciprocal 
if $u^{\deg P} P(t/u) \in F[t,u]$ is an alternating polynomial.

Let $P(t)$ be a self-reciprocal polynomial.
Then, by setting $u=1, t=\exp(x)$ in the above formula, we obtain
$P(\exp (x)) = \exp(dx) P (\exp (-x))$. This means that
$G_1 (x) = P (\exp(x))$ satisfies \eqref{eq:2.8} with $c=- \deg P$.
Similarly using skew self-reciprocal polynomials, we are able to construct
minus eigensequences.

The simplest examples of this construction are 
the binomials $P(t) = t^c  \pm 1$, which give
\[
G_1 (\boldsymbol{a}^{\pm},x )=
  \exp(cx) \pm 1 \in \mathscr{S}_1 (A_{-c})_1^{\pm}.
\]
\end{example}

\begin{example} \label{ex:3.6}
We can directly verify using \eqref{eq:1.2} that 
 the Bernoulli numbers $\boldsymbol{B}=\{B_n \}$  satisfies 
\[
 \boldsymbol{B} \in \mathscr{S} (A_1)_1^{+}.
\]
Similarly,
the Genocchi numbers $\boldsymbol{G}=\{G_n \}$ 
(\cite[Exercise 5.8 d]{MR1676282}) defined by 
\[
 G_1 (\boldsymbol{G},x) = \dfrac{2x}{\exp (x) +1}
\]
belongs to $\mathscr{S}_1(A_1)_1^{-}$.
\end{example}


\begin{example} \label{ex:3.7.5}
 Kummer's first formula for confluent hypergeometric function 
(there is a typo in \cite[(4.1.11)]{MR1688958})
\[
 F \left( \left.
\genfrac{}{}{0pt}{0}{\alpha  }{ \gamma } 
\right| x   \right) 
= \exp (x) \ F \left( \left.
\genfrac{}{}{0pt}{0}{\gamma - \alpha  }{ \gamma } 
\right| -x   \right) 
\]
leads to
\[
 F \left( \left.
\genfrac{}{}{0pt}{0}{\alpha  }{ 2 \alpha } 
\right| x   \right)  \in \mathscr{S}_1 (A_{-1})_1^{+}.
\]
\end{example}


We define polynomial sequences associated to a sequence.

\begin{definition} \label{def:3.8}
 For any $\boldsymbol{a} \in \mathscr{S}$, we define the associated
 polynomial sequence $ \boldsymbol{P}~ (\boldsymbol{a}, t)_k $ of weight $k$ by
\[
 P_n (\boldsymbol{a}, t)_k = [n] \boldsymbol{P}\&gt; (\boldsymbol{a}, t) 
= [n] (\boldsymbol{a}|_{[A_t]_k}) 
= (-1)^n\sum_{j=0}^n \binom{n+k-1}{j} a_{n-j} t^j \in F[t].
\]
\end{definition}

Z.-W. Sun \cite{MR1995582} and others adopted a slightly different
definition. In our notation, their polynomial is 
$ \left(\langle -1 \rangle \boldsymbol{a} \right)|_{[A_t A_0]_1} $.
Our definition seems to be more natural from our point of view.

The following proposition immediately follows from Proposition \ref{prop:2.12}.
\begin{proposition} \label{prop:3.9}
If $\boldsymbol{a} \in \mathscr{S} (A_c)_k^{s}$, then
$\boldsymbol{P}\&gt; (\boldsymbol{a}, t)_k  \in \mathscr{S} (A_{2t-c})_k^{s} $.
\end{proposition}

\begin{example} \label{ex:3.10}
A modified eigensequence $\langle -1 \rangle \boldsymbol{P}\&gt; 
(\boldsymbol{B}, t)_1 \in 
\mathscr{S} (A_{1-2t})_1^{+}$ of polynomials associated to the Bernoulli
sequence $\boldsymbol{B}$
is that of the Bernoulli polynomials.

The Euler polynomials $E_n (x)$ are defined by
\[
 \frac{2 e^{tx}}{e^x+1} = \sum_{n=0}^{\infty} E_n (t) \frac{x^n}{n!}.
\]
The Euler numbers $\boldsymbol{E}$ (see Example \ref{ex:2.5.1}) 
are related to the polynomials by 
$ E_n \left( \frac{1}{2} \right) = \frac{E_n}{(-2)^n} $.

It is easy to see that $\boldsymbol{E}'=\left\{ \frac{E_n}{(-2)^n} \right\} 
\in \mathscr{S} (A_0)_1^{+}$.
The associated polynomials to $\boldsymbol{E}'$ are given by
\[
 P_n (\boldsymbol{E}', t)_1 = (-1)^n E_n \left( t+ \frac{1}{2} \right)
\]
and it belongs to $\mathscr{S} (A_{2t})_1^{+}$.
\end{example}

The coefficients $\alpha_{2i}$ in \eqref{eq:3.7} are closely related to 
associated polynomials.
To show this relation, we first prove the following proposition,
which generalizes well-known symmetry formulas for 
Bernoulli and Euler polynomials.

\begin{proposition} \label{prop:3.10.3}
 If $\boldsymbol{a} \in \mathscr{S} (A_c)_k^{s}$, then we have
\[
 (-1)^n P_n (\boldsymbol{a}, c-t)_k = s P_n (\boldsymbol{a}, t)_k.
\]
\end{proposition}

\begin{proof}
By Proposition \ref{prop:3.9}, the sequence 
$\boldsymbol{P}\, (\boldsymbol{a},t)_k$ belongs to 
$\mathscr{S} (A_{2t-c})_k^{s}$. Therefore we have
$[n] \boldsymbol{P}\, (\boldsymbol{a},t)_k|_{[A_{2t-c}]_k} = 
s P_n (\boldsymbol{a},t)_k$.
On the other hand, using the definition of the associated polynomials, 
we compute
\begin{multline*}
[n] \boldsymbol{P}\, (\boldsymbol{a},t)_k|_{[A_{2t-c}]_k}
= [n]\left(\boldsymbol{a}|_{[A_{t}]_k} \right)|_{[A_{2t-c}]_k} 
= [n] \left(\boldsymbol{a}|_{[A_{t}A_{2t-c}]_k} \right) \\
=[n] \left(\boldsymbol{a}|_{[A_{c-t}A_{0}]_k} \right)
=[n] \left(\boldsymbol{P}\, (\boldsymbol{a},c-t) \right)_k|_{[A_0]_k}
= (-1)^n P_n (\boldsymbol{a}, c-t)_k.
\end{multline*}
Here we used a similar method as in Proposition \ref{prop:2.12}.
This completes the proof.
\end{proof}

Now we can show the following proposition.

\begin{proposition} \label{prop:3.10.5}
Let $\boldsymbol{a} \in \mathscr{S} (A_c)_k^{\pm}$.
We have an explicit formula for \eqref{eq:3.7} and \eqref{eq:3.8}:
\[
 G_0 (\boldsymbol{a},x) = \sum_{i = 0}^{\infty}
P_i \left( \boldsymbol{a}, \frac{c}{2} \right)_k  x^i \mathscr{E}_{k+i} (x).
\]
\end{proposition}
\begin{proof}
First note that the central value $t=c/2$ is a root of 
$P_n (\boldsymbol{a},t)$ for odd $n$ 
if $\boldsymbol{a} \in \mathscr{S} (A_c)_k^{+} $ and 
for even $n$ if  $\boldsymbol{a} \in \mathscr{S} (A_c)_k^{-} $.
This fact readily follows from 
Proposition \ref{prop:3.10.3}.
We calculate the coefficient of $x^m$ in the 
right hand side of the equality using \eqref{eq:2.2}:
\[
 \sum_{j=0}^{m} \binom{m+k-1}{j} P_{m-j} 
\left( \boldsymbol{a},\frac{c}{2} \right)_k \left( -\frac{c}{2} \right)^j
=[m] \left((\boldsymbol{a}|_{[A_{c/2}]_k})|_{[A_{c/2}]_k}) \right).
\]
Since $[A_{c/2}]_k$ is an involution, this is equal to 
$[m] \boldsymbol{a}$. This proves the proposition.
\end{proof}

\begin{example}
As an example, we compute the explicit expansion of the generating 
function $G_0 (\boldsymbol{F},x) \in \mathscr{S}_0 (A_{-1})_1^{-}$ 
of Fibonacci numbers.
It is easy to compute 
\[
\sum_{n \ge 0} P_n \left( \boldsymbol{F}, -\frac{1}{2} \right)_1 x^n
= G_0|_{[A_{-1/2}]_1} (\boldsymbol{F},x) = \frac{-x}{1-\frac{5}{4}x^2}.
\] 
Therefore we get
\[
G_0 (\boldsymbol{F},x) =\sum_{i = 0}^{\infty} \left(\frac{5}{4}\right)^{i} 
x^{2i+1} \mathscr{E}_{2i+2} (x).
\]
Taking $[x^n]$ of the both sides, we obtain a classical formula
due to Catalan:
\[
 F_n = \frac{1}{2^{n-1}} \sum_{i=0}^{\lfloor \frac{n-1}{2}\rfloor} 
5^i \  \binom{n}{n-2i-1}.
\]
\end{example}


\begin{example} \label{ex:3.11}
There is another important family of polynomial eigensequences.
They are orthogonal polynomials. 
The Jacobi polynomial $P_n^{(\alpha, \beta)} (t)$ 
is defined in terms of Gaussian hypergeometric 
function (\cite[Definition 2.5.1]{MR1688958}) and 
its exponential generating function is 
\[
 \frac{2^{\alpha +\beta }}{R (1-x+R)^{\alpha} (1+x+R)^{\beta }}
\]
(see \cite[Theorem 6.4.2]{MR1688958})
where $R$ is defined by
\[
 R=\sqrt{1-2tx+x^2},
\]
which belongs to $\mathscr{S}_0 (A_{-2t})_{-1}^{+}$
by Examples \ref{ex:3.2} and \ref{ex:3.4}.
The product $(1-x+R)(1+x-R) = 2 (R+(1-xt))$ is also in 
$\mathscr{S}_0 (A_{-2t})_{-1}^{+}$.
Thus if $\alpha = \beta $, then the exponential generating function
belongs to $\mathscr{S}_0 (A_{-2t})_{\alpha+1}^{+}$.
We conclude 
\[
\left\{ \frac{P_n^{(\alpha, \alpha)} (t )}{n!} \right\}_{n \ge 0}
\in \mathscr{S} (A_{-2t})_{\alpha+1}^{+}.
\]
When $\alpha =\beta =0 $, the polynomial $P_n (t) = P_n^{(0,0)}(t)$ is called 
the Legendre polynomial.

The following orthogonal polynomials also form polynomial
 eigensequences:
\begin{itemize}
 \item The Gegenbauer polynomial $C_n^{\lambda } (t) $ 
(\cite[p.302]{MR1688958}).
 \[
    G_0 (\{ C_n^{\lambda} (t) \}, x ) = \frac{1}{R^{2\lambda}} \in 
\mathscr{S}_0 (A_{-2t})_{2 \lambda}^{+}.
\]
The special case $U_n (t) = C_n^1 (t)$ is called Chebyshev polynomial of 
second kind.
 \item The Chebyshev polynomial of first kind $T_n (t)$ 
(\cite[Remark 2.5.3]{MR1688958}).
The ordinary generating function is
\[
 \frac{1-tx}{R^2} \in \mathscr{S}_0 (A_{-2t})_{1}^{+}.
\]
\item The Hermite polynomial $H_n(t)$ (\cite[Section 6.1]{MR1688958}).
The exponential generating function is 
$\exp (2tx-t^2) \in \mathscr{S}_1 (A_{-4t})_{1}^{+} $.
\end{itemize}
\end{example}

\section{Differential operations on eigensequences} \label{sec:4}
The main result in this section is the following 
analogue of the Rankin-Cohen differential operator.
Since the sign differs from the original formula 
(see \cite[Section 1]{MR1280058}), 
we state and prove the following theorem for a general 
lower triangular matrix 
$A=\begin{bmatrix}
    a &amp; 0 \\ c &amp; d
   \end{bmatrix}$.

\begin{theorem} \label{thm:4.1}
Let $n$ be a non-negative integer and $k$ and $\ell$ integers 
 such that the following conditions are \textit{not} satisfied:
\begin{equation} \label{eq:4.1}
-n &lt; k \le 0
\ \text{   or   }
-n &lt; \ell \le 0.
\end{equation}
For $f \in \mathscr{S}_0 (A)^{s_1}_k $ and 
$g  \in \mathscr{S}_0 (A)^{s_2}_{\ell } $, we define the $n$-th
Rankin-Cohen bracket of $f$ and $g$ by the formula
\[
 [f,g]_{n} (x) = \sum_{r+s=n} (-1)^r \binom{n+k-1}{s} \binom{n+\ell -1}{r}
f^{(r)} (x) g^{(s)} (x).
\]
Then we have
\[
 [f,g]_n \in \mathscr{S}_0 (A)_{k+\ell + 2n}^{\frac{s_1 s_2}{(ad)^n}}.
\]
\end{theorem}
\begin{proof}
Our formulation and proof are based on Zagier's article \cite{MR1280058}.
Therefore we only give a sketch of the proof.
However we have to take care of negative $k$ and $\ell $.
We first associate a formal power series $\tilde{f} (x, t)$ to $f(x)$ 
\[ 
 \tilde{f} (x,t) =    \sum_{i=0}^{n} 
\frac{f^{(i)}(x)}{(i+k-1)^{\underline{i}}} 
\frac{t^i}{i!}.
\]
The formal power series $ \tilde{g} (x,t)$ for $g(x)$ is similarly defined.
This definition makes sense if \eqref{eq:4.1} are not satisfied.

Then we can prove
\[
[t^n]  \tilde{f} (x,-t)  \tilde{g} (x,t)
= \frac{[f,g]_n (x)}
{(n+k-1)^{\underline{n}}(n+\ell -1)^{\underline{n}} }
\]
On the other hand, we can show the following formulas
for the higher derivatives
by induction:
\begin{align*}
 \frac{(ad)^n f^{(n)} \left( \frac{ax}{cx+d} \right)}
{n! (n+k-1)^{\underline{n}}}
&amp; = s_1 \sum_{m=0}^n 
\frac{c^{n-m} (cx+d)^{k+n+m}}{(n-m)!} 
\frac{f^{(m)}(x)}{m! (m+k-1)^{\underline{m}}}.
\end{align*}
It follows from these formulas that $\tilde{f}$ satisfies the 
transformation law
\[
 \tilde{f} \left( \frac{ax}{cx+d}, \frac{(ad)t}{(cx+d)^2} \right)
\equiv s_1 (cx +d)^k \exp \left( \frac{ct}{cx+d} \right) \tilde{f} (x,t)
\pmod{t^{n+1}}.
\]
A similar formula holds also for $\tilde{g}$:
\[
 \tilde{g} \left( \frac{ax}{cx+d}, \frac{(ad)t}{(cx+d)^2} \right)
\equiv s_2 (cx +d)^{\ell } \exp \left( \frac{ct}{cx+d} \right) \tilde{g} (x,t)
\pmod{t^{n+1}}.
\]
Multiplying the both sides, we have
\[
 \tilde{f} \left( \frac{ax}{cx+d}, -\frac{(ad)t}{(cx+d)^2} \right)
 \tilde{g} \left( \frac{ax}{cx+d}, \frac{(ad)t}{(cx+d)^2} \right)
\equiv s_1 s_2 (cx +d)^{k+ \ell } \tilde{f}(x,-t) \tilde{g} (x,t)
\pmod{t^{n+1}}.
\]
Comparing the coefficients of $t^n$, we obtain
\[
 [f,g]_n \left( \frac{ax}{cx+d} \right) = \frac{s_1 s_2}{(ad)^n} 
(cx+d)^{k+\ell + 2n} [f,g]_n (x).
\]
This means that 
$ [f,g]_n \in \mathscr{S}_0 (A)_{k+\ell + 2n}^{\frac{s_1 s_2}{(ad)^n}}$.
This completes the proof of the theorem.
\end{proof}

As in the paper \cite{MR1280058}, the following identities involving 
the brackets hold for $f \in \mathscr{S}_0 (A)^{s_1}_{k },
g \in \mathscr{S}_0 (A)^{s_2}_{\ell }$ and 
$h \in \mathscr{S}_0 (A)^{s_3}_{m }$:
\begin{gather*}
 [f,g]_n = (-1)^n [g,f]_n, \\
  [[f,g]_1, h]_1 + [[g,h]_1, f]_1 + [[h,f]_1, g]_1 =0 \quad \text{(the Jacobi 
identity)}, \\
  [[f,g]_0, h]_1 + [[g,h]_0, f]_1 + [[h,f]_0, g]_1 =0 , \\
 m [[f,g]_1, h]_0 + k [[g,h]_1, f]_0 + \ell [[h,f]_1, g]_0 =0 , \\
  [[f,g]_0, h]_1 = [[g,h]_1, f]_0 + [[h,f]_1, g]_0  , \\
(k + m + \ell ) [[f,g]_1,h]_0 = k[[h,f]_0, g]_1 -\ell [[g,h]_0,f]_1.
\end{gather*}

Here we write down the brackets of low degree in the case of $A=A_c$:
\begin{align*}
 [f,g]_0 &amp; = fg \in \mathscr{S}_0 (A_c)^{s_1 s_2}_{k+ \ell } , \\
 [f,g]_1  &amp; = k f g' - \ell f' g
\in \mathscr{S}_0 (A_c)^{- s_1 s_2}_{k+ \ell +2 }, \\
 [f,g]_2 &amp; = \frac{\ell (\ell +1)}{2} f'' g -(k+1)(\ell +1 ) f'g'
+\frac{k(k+1)}{2} f g'' \in \mathscr{S}_0 (A_c)^{s_1 s_2}_{k+ \ell +4 }.
\end{align*}

\begin{example} \label{ex:4.3}
Let $\boldsymbol{a} = \{a_i \} \in \mathscr{S}(A_c)^{s}_k \ (k &gt;0)$.
Since $x^{\ell } \in \mathscr{S}(A_c)^{(-1)^{\ell}}_{-\ell} $ 
(see \eqref{eq:3.0}), we have
\begin{align*}
[G_0 (\boldsymbol{a} ,x), x^{\ell }]_n &amp;=  
\sum_{s=0}^{n} (-1)^{n-s} \binom{n+k-1}{s} \binom{n-\ell-1}{n-s} 
\left( \sum_{i=0}^{\infty} (i+n-s)^{\underline{n-s} } a_{i+n-s} x^i \right)
 \ \ell^{\underline{s}} \ x^{\ell -s} \\
&amp;= \sum_{i=0}^{\infty} \left(\sum_{s=0}^n (-1)^{n-s} \binom{n+k-1}{s}
\binom{n-\ell -1}{n-s} \ell^{\underline{s}} (i+n-\ell)^{\underline{n-s}} 
\right) a_{i+n-\ell } x^i \\
&amp;= \sum_{i=0}^{\infty} \left(\sum_{s=0}^n (-1)^{n-s} \binom{n+k-1}{s}
(-1)^{n-s} \binom{\ell-s}{n-s} s! \binom{\ell}{s} (n-s)! 
\binom{i-\ell +n}{n-s} \right) a_{i+n-\ell } x^i \\
&amp;=n! \ \binom{\ell}{n} 
\sum_{i=0}^{\infty} \left( \sum_{s=0}^n \binom{n+k-1}{s} \binom{i-\ell+n}{n-s}
a_{i+n-\ell } \right) x^i \\
&amp;= n! \ 
\binom{\ell}{n} \sum_{i=0}^{\infty} 
 \binom{2n+k+i-\ell -1}{n} a_{i+n-\ell } x^i 
\in \mathscr{S}_0 (A_c)^{s (-1)^{n+\ell} }_{k- \ell +2n }.
\end{align*}
for $n \le \ell$.
Here the last equality follows from Vandermonde's convolution.
An interesting case is $\ell =2n$. Then $[*,x^{2n}]_n$ preserves
the weight and we have
\[
 [G_0 (\boldsymbol{a} ,x), x^{2n }]_n 
= (2n)^{\underline{n}} \sum_{i=n}^{\infty}
\binom{i+k-1}{n} \, a_{i-n} x^i
\in \mathscr{S}_0 (A_c)^{(-1)^n s }_{k }.
\]
We conclude that
\[
\left\{ \binom{i+k-1}{n} \, a_{i-n} \right\}_{i \ge 0}  
\in \mathscr{S}(A_c)^{(-1)^n s}_k \]
 and 
\[
 \{ (i+k-1)^{\underline{n}} \ a_{i-n} \}_{i \ge 0}
 \in \mathscr{S}(A_c)^{(-1)^n s}_k .
\]
\end{example}

There is also an analogue of Theorem \ref{thm:4.1} for exponential 
generating functions.

\begin{theorem} \label{thm:4.8}
Let $n$ be a non-negative integer.
 If $f \in \mathscr{S}_1(A_{c_1})_1^{s_1}$ and 
$g \in \mathscr{S}_1 (A_{c_2})_1^{s_2}$, then
\[
 \{f,g\}_n = \sum_{j=0}^n \binom{n}{j} 
c_2^j (-c_1)^{n-j} f^{(j)}(x) g^{(n-j)}(x) 
\in \mathscr{S}_1 (A_{c_1+ c_2})_1^{(-1)^n s_1 s_2}.
\]
\end{theorem}

\begin{proof}
Using Leibniz's rule, we calculate
\begin{align*}
&amp; \exp (-(c_1 +c_2)x ) \sum_{j=0}^n \binom{n}{j} c_2^j (-c_1)^{n-j}
 f^{(j)}(-x) \otimes g^{(n-j)} (-x) \\
&amp; =\exp (-(c_1 +c_2)x ) \sum_{j=0}^n \binom{n}{j} c_2^j (-c_1)^{n-j} \\
&amp;\times \left( s_1 \exp (c_1 x) \left(  -c_1 -\frac{d}{dx} \right)^j f(x)
\right)
\otimes 
\left( s_2 \exp (c_2 x) \left(  -c_2 -\frac{d}{dx} \right)^{n-j} g(x)
\right)\\
&amp;= s_1 s_2 \sum_{j=0}^n \binom{n}{j} c_2^j (-c_1)^{n-j} 
\left( \left(-c_1-\frac{d}{dx} \right)^j f(x) \otimes
\left(-c_2-\frac{d}{dx} \right)^{n-j} g(x) 
\right) \\
&amp;= s_1 s_2 \left( c_2 \left( -c_1 -\frac{d}{dx}\right)
+ (-c_1) \left( -c_2 -\frac{d}{dx}\right)
\right)^n f(x) \otimes g(x) \\
&amp;=(-1)^n s_1 s_2 \left( c_2 \frac{d}{dx} -c_1 \frac{d}{dx} \right)^n
f(x) \otimes g(x) \\
&amp;= (-1)^n s_1 s_2 \sum_{j=0}^n \binom{n}{j} 
c_2^j (-c_1)^{n-j} f^{(j)}(x) g^{(n-j)}(x).
\end{align*}
This implies that
\[
 \exp(-(c_1+c_2)x) \{ f,g \}_n (-x) = (-1)^n s_1 s_2 \{ f,g\}_n (x).
\]
The result follows from \eqref{eq:2.8}.
\end{proof}

If $f(x)=G_1 (\boldsymbol{a},x)$ and $g(x) = G_1 (\boldsymbol{b},x)$,
then we compute
\begin{equation} \label{eq:4.8}
 \left[ \frac{x^m}{m!}\right] \{ G_1 (\boldsymbol{a},x), G_1
  (\boldsymbol{b},x) \}_n
= \sum_{j=0}^n \binom{n}{j} \ c_2^j (-c_1)^{n-j} 
\sum_{i=0}^{m}  \binom{m}{i} \ a_{i+j} b_{m+n-i-j}.
\end{equation}
In particular, if $c_2=-c_1$, then
\[
 \left[ \frac{x^m}{m!}\right] 
\{ G_1 (\boldsymbol{a},x), G_1
  (\boldsymbol{b},x) \}_n = (-c_1)^n 
 \left[ \frac{x^{m+n}}{(m+n)!}\right] 
\{ G_1 (\boldsymbol{a},x), G_1
  (\boldsymbol{b},x) \}_0.
\]

We again write down the brackets of low degree:
\begin{align*}
 \{f,g\}_0 &amp;= fg \in \mathscr{S}_1 (A_{c_1+ c_2})_1^{s_1 s_2}, \\
 \{f,g\}_1 &amp;= -c_1 f g' +c_2 f' g 
\in \mathscr{S}_1 (A_{c_1+ c_2})_1^{-s_1 s_2}, \\
 \{f,g\}_2 &amp;= c_1^2 fg''-2c_1 c_2 f'g' +c_2^2 f'' g 
 \in \mathscr{S}_1 (A_{c_1+ c_2})_1^{s_1 s_2}. 
\end{align*}

These brackets also satisfy the analogous properties of the Rankin-Cohen 
brackets. We omit to write down these formulas.

We conclude this section with a few remarks.

We can apply Theorem \ref{thm:4.8} only to
eigensequences of weight $1$. Thus it is worth considering
a method to transform a general eigensequence to a weight-$1$ sequence.

The simplest method is shifting. Namely
by shifting an eigensequence of positive weight to the right, 
it eventually becomes a sequence of weight $1$.
The right shifts correspond to iterated formal integration 
of the exponential generating function (cf. Proposition \ref{prop:2.2}).  

There is another interesting method.
Let $\boldsymbol{a} \in \mathscr{S} (A_c)_k^{\pm} \ (c \neq 0)$.
We choose integer parameters $\alpha$ and $\beta$ and define a new sequence
$\boldsymbol{b}$ by
\begin{equation} \label{eq:4.11}
 G_1 (\boldsymbol{b},x) = \exp ( \beta x)
G_0 \left( \boldsymbol{a}, \frac{1}{c} (\exp (\alpha x) -1) \right).
\end{equation}
An easy calculation shows that
\begin{equation} \label{eq:4.12}
  G_1 (\boldsymbol{b},x) = \pm \exp ((2 \beta - k \alpha ) x )
  G_1 (\boldsymbol{b},-x)
\end{equation}
holds.
By \eqref{eq:2.8} we conclude
$\boldsymbol{b} \in \mathscr{S} (A_{k\alpha -2\beta})_1^{\pm}$.

The transformation \eqref{eq:4.11} is closely related to a generalization
of the Akiyama-Tanigawa algorithm (see \cite{MR1891413} and \cite{MR1800883}). 
We explain this connection briefly.
Starting from the initial sequence $\boldsymbol{a} =\{ a_{0,m} \}_{m \ge 0}$, 
we define new sequences $\{ a_{n,m} \}_{m \ge 0}$ for $n=1,2,\dots$ by
\begin{equation} \label{eq:4.13}
 a_{n,m} = (\alpha m + \beta ) a_{n-1,m} + (\gamma m+ \delta ) a_{n-1, m+1}
\end{equation}
with given parameters $\alpha, \beta$ , $\gamma $ and $\delta$.
Then we collect the $0$-th terms and form a sequence 
$\boldsymbol{b}= \{ a_{n,0} \}_{n \ge 0}$.
If $\delta = \gamma $, 
then by a similar method as \cite{MR1848943}, we can prove
\[
 a_{n,0} = \sum_{j=0}^n \binom{n}{j} \beta^{n-j} \left( \sum_{m=0}^n
\genfrac{\{}{\}}{0pt}{0}{n}{m} \alpha^{n-m} \gamma^m m! \, a_{0,m} \right),
\]
where $\genfrac{\{}{\}}{0pt}{0}{n}{m}$ is the Stirling number 
of the second kind and 
\[
 G_1 (\boldsymbol{b}, x) = \exp (\beta x) G_0 \left( \boldsymbol{a}, 
\frac{\gamma}{\alpha} (\exp (\alpha x) -1 )\right).
\]
Chen \cite{MR1848943} proved these formulas for the special cases where 
$(\alpha, \beta , \gamma ) = (1,1,-1)$ and $(1,0,-1)$.

We have chosen $\gamma =  \alpha /c$ in \eqref{eq:4.13}.

Incidentally, if $\alpha = \gamma =0$, 
then the sequences $\{ a_{n,m} \}$
form a matrix called generalized Seidel matrix.
In this case, assuming $\delta \neq 0$, we have
\[
 G_1 (\boldsymbol{a},x) = \exp \left( -\frac{\beta x}{\delta} \right)
G_1 \left( \boldsymbol{b}, \frac{x}{\delta} \right)
\]
(see \cite[Theorem 2.1]{MR2320995}).
Hence this transformation can be written by our action 
(but it is not a convolution).
Moreover, if  $\boldsymbol{a} \in \mathscr{S} (A_c)_1^{s}$, then 
we can prove $\boldsymbol{b} \in \mathscr{S} (A_{c\delta -2 \beta })_1^{s}$.

\section{Involutive action on endomorphisms} \label{sec:5.5}
In this section, we study the action of the involutions on the 
endomorphism algebra $ \mathrm{End} (\mathscr{S})$ of the vector space
$\mathscr{S}$.

\begin{definition} \label{def:5.5.1}
Let $A \in \mathrm{GL}_2 (F)$ be a lower triangular regular matrix 
of order $2$.
We define a left action of $A$ of weight $k$ on 
 $ \mathrm{End} (\mathscr{S})$ by
\begin{equation} \label{eq:5.5.1}
( [A]_k f ) (\boldsymbol{a}) = (f(\boldsymbol{a}|_{[A]_k}))|_{[A]_k}
\end{equation}
for $f \in \mathrm{End}(\mathscr{S}) $ and 
$\boldsymbol{a} \in \mathscr{S}$. 
\end{definition}
Since $A$ acts as an involution on $ \mathrm{End} (\mathscr{S})$,
we can decompose $\mathrm{End} (\mathscr{S})$ as
an $F[A]_k$-module:
\[
 \mathrm{End} (\mathscr{S}) =  \mathrm{End} (\mathscr{S})^{+} \oplus 
 \mathrm{End} (\mathscr{S})^{-},
\]
where the eigenspaces are given by
\[
  \mathrm{End} (\mathscr{S})^{+} 
= \{ f \in  \mathrm{End} (\mathscr{S}) \ | \ [A]_k f = f \}, \quad
 \mathrm{End} (\mathscr{S})^{-}
= \{ f \in  \mathrm{End} (\mathscr{S}) \ | \ [A]_k f = -f \}.
\]
It is easy to see that $  \mathrm{End} (\mathscr{S})^{+} $ 
coincides with the submodule $  \mathrm{End}_{F[A]_k} (\mathscr{S}) $ 
of $F[A]_k$-endomorphisms. 
From this fact, we can show the following proposition easily.
\begin{proposition} \label{prop:5.5.2}
If $\boldsymbol{a}^{+} \in \mathscr{S}(A)_k^{+}, 
\boldsymbol{a}^{-} \in \mathscr{S}(A)_k^{-}$,
$f^{+} \in \mathrm{End} (\mathscr{S})^{+} and
f^{-} \in \mathrm{End} (\mathscr{S})^{-} $, then we have
\[
 f^{+} (\boldsymbol{a}^{+}) \in \mathscr{S}(A)_k^{+}, \quad
 f^{+} (\boldsymbol{a}^{-}) \in \mathscr{S}(A)_k^{-}, \quad
 f^{-} (\boldsymbol{a}^{+}) \in \mathscr{S}(A)_k^{-}, \quad
 f^{-} (\boldsymbol{a}^{-}) \in \mathscr{S}(A)_k^{+}. \]
\end{proposition}
This proposition gives
 another method to generate new eigensequences.


We shall compute the action of $[A_c]_k $ on $f \in \mathrm{End} (\mathscr{S})$
explicitly for several elementary 
endomorphisms $f$.

In the following, until the end of this section,
we restrict ourselves to the case where the weight $k=1$, for simplicity.

\begin{proposition} \label{prop:5.5.7}
 Let $\partial^m$ be the differentiation on exponential generating function:
\[
 \partial^m : G_1 (\boldsymbol{a}, x) \mapsto \frac{d^m}{dx^m}
G_1 (\boldsymbol{a},x).
\]
Then we have
\begin{equation} \label{eq:5.5.2}
[A_c]_1 \partial^m (G_1 (\boldsymbol{a}, x)) = 
 \left(-c- \partial \right)^m G_1 (\boldsymbol{a},x). 
\end{equation}
\end{proposition}
\begin{proof}
Applying Leibniz's rule to \eqref{eq:2.5}, we have
\begin{align*}
 \partial^m   G_1 (\boldsymbol{a}|_{[A_c]_1},x) &amp;= \sum_{i=0}^{m} 
\binom{m}{i} (-c)^{m-i} \exp (-cx) \ (-1)^i \partial^i \, G_1 (\boldsymbol{a},-x)\\
&amp;= (-1)^m \sum_{i=0}^{m} 
\binom{m}{i}\  c^{m-i} \exp (-cx) \  \partial^i \, G_1 (\boldsymbol{a},-x)
\end{align*}
Therefore we get
\begin{align*}
[A_c]_1 \partial^m (G_1 (\boldsymbol{a}, x)) &amp; = \exp (-cx) (-1)^m
\sum_{i=0}^m \binom{m}{i} c^{m-i} \exp (cx) \ \partial^i\,  G_1 (\boldsymbol{a},x) \\
&amp; = (-1)^m\sum_{i=0}^m \binom{m}{i} c^{m-i} \partial^i\  G_1 (\boldsymbol{a},x) \\
&amp;= (-c-\partial)^m G_1 (\boldsymbol{a},x).
\end{align*}
\end{proof}

We rewrite this result in terms of sequence.

\begin{proposition} \label{prop:5.5.4}
Let $\boldsymbol{a} \in \mathscr{S}$ and 
 $f=L^m$ the map shifting $m \ (m \ge 1)$ terms to the left:
\[
 [n](L^m (\boldsymbol{a})) = a_{m+n}, \quad [0](L^m (\boldsymbol{a})) = a_{m}.
\]
Then we have
\begin{align*}
   [n] (([A_c]_1 L^m) (\boldsymbol{a})) &amp; = 
(-1)^m \sum_{i=0}^m 
 \binom{m}{i} c^{m-i} a_{n+i} = [m] (L^n (\boldsymbol{a})|_{[A_c]_1}), \\
[0] (([A_c]_1 L^m) (\boldsymbol{a})) &amp; = (-1)^m \sum_{i=0}^m
 \binom{m}{i} c^{m-i} a_{i} = [m] (\boldsymbol{a}|_{[A_c]_1}).
\end{align*}
\end{proposition}
\begin{proof}
We obviously have 
$\partial^m G_1 (\boldsymbol{a},x) = G_1 (L^m \boldsymbol{a}, x)$.
Hence the result follows by taking the $n$-th terms of \eqref{eq:5.5.2}.
\end{proof}

\begin{proposition} \label{prop:5.5.5}
 Let $f=\Delta$ be the difference operator defined by 
$a_n \mapsto a_{n+1}-a_n$. Then we have
\begin{align*}
 [n]( ([A_c]_1 \Delta^m ) (\boldsymbol{a}) ) 
&amp; = (-1)^m \sum_{j=0}^m \binom{m}{j} (1+c)^{m-j} a_{n+j} 
= [m] (L^n (\boldsymbol{a})|_{[A_{1+c}]_1}), \\
 [0]( ([A_c]_1 \Delta^m ) (\boldsymbol{a}) ) &amp; =
 (-1)^m \sum_{j=0}^m \binom{m}{j} (1+c)^{m-j} a_{j} 
=[m] (\boldsymbol{a}|_{[A_{1+c}]_1}).
\end{align*}
\end{proposition}
\begin{proof}
 It is easy to show that
\[
 [n] (\Delta^m (\boldsymbol{a})) = \sum_{i=0}^m \binom{m}{i} (-1)^{m+i} a_{n+i}.
\]
Hence we have
\[
 \Delta^m (G_0 (\boldsymbol{a}|_{[A_c]_1},x)) 
= \sum_{i=0}^m \binom{m}{i} (-1)^{m+i} 
 L^i G_0 (\boldsymbol{a}|_{[A_c]_1},x) ).
\]
From the definition \eqref{eq:5.5.1} it follows  that
\[
 ([A_c]_1 \Delta^m ) (G_0 (\boldsymbol{a},x)) 
= \sum_{i=0}^m \binom{m}{i} (-1)^{m+i} 
([A_c]_1 L^i) G_0 (\boldsymbol{a},x).
\]
Therefore, by Proposition \ref{prop:5.5.4}, the $n$-th term is given by
\begin{align*}
[n]( ([A_c]_1 \Delta^m ) (\boldsymbol{a}) ) &amp;
=(-1)^m\sum_{i=0}^m \sum_{t=0}^i  \binom{m}{i} \binom{i}{t} c^t
 a_{n+i-t} \\
&amp;= (-1)^m\sum_{j=0}^m \sum_{i=j}^{m}  \binom{m}{i} \binom{i}{j} 
c^{i-j} a_{n+j}.
\end{align*}
The binomial identity 
$ \displaystyle
 \binom{m}{i} \binom{i}{j} = \binom{m}{j}\binom{m-j}{m-i}
$
yields
\[
 [n]( ([A_c]_1 \Delta^m ) (\boldsymbol{a}) ) 
 = (-1)^m 
\sum_{j=0}^m \binom{m}{j}  \sum_{i=0}^{m-j} \binom{m-j}{i} c^i a_{n+j} 
= (-1)^m \sum_{j=0}^m \binom{m}{j} (1+c)^{m-j} a_{n+j} 
\]
as desired.
\end{proof}

Another differential operators producing new eigensequences are
given in the following corollary,
whose proof readily follows from Propositions \ref{prop:5.5.2}
and \ref{prop:5.5.7}.

\begin{corollary}
Let $G (x) \in \mathscr{S}_1 (A_{c})_1^{t}$.
For any polynomial $p (\partial ) \in F[\partial ]$, we define 
 \[
  p ( \partial ) G (x) = p \left(\frac{d}{dx} \right) G (x) .
 \]
If $p ( \partial ) \in F[ \partial ]$ satisfies
\begin{equation} \label{eq:4.2}
 p (\partial )= s \cdot  p (-c -\partial)
\end{equation}
with some $s \in \{\pm 1\}$,
then we have
 \[
 p (\partial ) G (x) \in \mathscr{S} (A_{c})_1^{st}.
\]
\end{corollary}

The following lemma tells us how to find polynomials
satisfying \eqref{eq:4.2}.

\begin{lemma} \label{lem:4.5}
 Let $u (\partial )$ be an even (resp. odd) polynomial in $\partial $.
Then $p (\partial ) = u \left( \partial +\dfrac{c}{2} \right)$ 
satisfies \eqref{eq:4.2}
with $s =1$ (resp. $s=-1$). Conversely every polynomial satisfying 
\eqref{eq:4.2} is obtained in this way.
\end{lemma}
\begin{proof} 
The first half follows from an easy calculation.
We shall prove the latter half.
Notice that the map
 $\partial \mapsto -c-\partial $ is an involution on $F[\partial]$. 
Therefore any polynomial 
$p$ satisfying \eqref{eq:4.2} is obtained by 
$p (\partial)= h(\partial ) + s \ h(-c-\partial )$ for some 
$h(\partial ) \in F[\partial ]$.
Then we have
\[
 p \left( \partial -  \dfrac{c}{2} \right) =
h \left(\partial - \dfrac{c}{2} \right) 
+ s \ h \left(- \partial - \dfrac{c}{2} \right).
\] 
This shows that 
$p \left( \partial - \dfrac{c}{2} \right)$ is an even (resp. odd) polynomial
if $s=1$ (resp. $s=-1$). 
\end{proof}

\begin{example} \label{ex:4.6}
We give some examples of polynomials satisfying \eqref{eq:4.2}.
First we use $u (\partial )$ in Lemma \ref{lem:4.5} to obtain
\begin{align}
&amp; \frac{c}{2}   + \partial \quad ( s =-1), \label{eq:4.4} \\
&amp;  c \partial  + \partial^2  \quad  (s = 1) \label{eq:4.5} . 
\end{align}
Although we have $  \left( \partial + \dfrac{c}{2} \right)^2 =
\partial^2 + c \partial +\dfrac{c^2}{4}$,
we can drop the constant term since
$1 \in F [\partial ] $ defines an operator with $s=1$.
These two operators are found in \cite[Corollary 3.1]{MR1851531}
as operators related to $[A_{-1}]_1$.

We have another type of operators like
\begin{align}
&amp; \partial^n \pm \left( -c -\partial \right)^n , \label{eq:4.6} \\
&amp; \partial^n (-c-\partial )^m \pm (-c-\partial )^n \partial^m
 \label{eq:4.7}, \\
&amp; (\partial^2 -c^2)^n \pm (\partial^2 +2c \partial)^n \label{eq:4.7.5}
\end{align}
with $ s = \pm 1$ and $m,n \in \Z_{&gt;0}$.
\end{example}

These operators will be used to produce identities of eigensequences
in the next section.

\section{Identities for eigensequences} \label{sec:5}
In this section, we shall deduce some identities involving eigensequences
as an application of the theory we have developed so far 
in the preceding sections. 

First we show that
 we can construct eigensequences from arbitrary
sequences easily by applying `twisted' endomorphisms.

\begin{proposition} \label{prop:2.1}
 Let $f \in \mathrm{End} (\mathscr{S})$ and 
$\boldsymbol{a} \in \mathscr{S}$.
Let $A \in \mathrm{GL}_2 (F)$ be a lower triangular matrix 
of order $2$.
Then we have
\begin{enumerate}
 \item \label{2.1.1} 
$f(\boldsymbol{a}) + ([A]_k f) (\boldsymbol{a}|_{[A]_k}) \in
       \mathscr{S}(A)_k^{+}$;
 \item \label{2.1.2} $f(\boldsymbol{a}) - 
                   ([A]_k f) (\boldsymbol{a}|_{[A]_k}) 
              \in \mathscr{S}(A)_k^{-}$;
\item  \label{2.1.3} $([A]_k f )(\boldsymbol{a}) + 
f (\boldsymbol{a}|_{[A]_k}) \in \mathscr{S}(A)_k^{+}$;
 \item \label{2.1.4} $([A]_k f ) (\boldsymbol{a}) - 
f (\boldsymbol{a}|_{[A]_k}) \in 
\mathscr{S}(A)_k^{-}$.
\end{enumerate}
\end{proposition}
\begin{proof}
The assertions \ref{2.1.3} and \ref{2.1.4} follow from \ref{2.1.1} and
\ref{2.1.2} respectively taking $ [A]_k f$ as $f$.
Therefore we have only to prove the first two assertions.
Recall that we have projections $\pi (A)_k^{+} $ (resp. $\pi (A)_k^{-}$)
from $\mathscr{S}$ to $\mathscr{S}(A)_k^{+}$ (resp. $\mathscr{S}(A)_k^{+}$)
(see the remark after Proposition \ref{prop:2.12}).
By Definition \ref{def:5.5.1}, we have
\[
 f(\boldsymbol{a}) + ([A]_k f) (\boldsymbol{a}|_{[A]_k}) =
f(\boldsymbol{a}) + f (\boldsymbol{a})|_{[A]_k}
= 2 \pi (A)_k^{+} (f(\boldsymbol{a})) \in \mathscr{S}(A)_k^{+}.
\]
Similarly, we get
\[
 f(\boldsymbol{a}) - ([A]_k f) (\boldsymbol{a}|_{[A]_k}) =
2 \pi (A)_k^{-} (f (\boldsymbol{a})) \in \mathscr{S}(A)_k^{-}.
\]
This completes the proof.
\end{proof}

We have the following corollary.

\begin{corollary} \label{cor:2.3}
 Let $f, g \in \mathrm{End}(\mathscr{S})$ and $\boldsymbol{a} \in \mathscr{S}$.
Then we have 
\begin{enumerate}
 \item $([A]_k f \circ g) (\boldsymbol{a}) 
+ ( f \circ [A]_k g) (\boldsymbol{a}|_{[A]_k}) 
\in \mathscr{S}(A)_k^{+}$;
\item $([A]_k f \circ g) (\boldsymbol{a}) 
- ( f \circ [A]_k g) (\boldsymbol{a}|_{[A]_k}) 
\in \mathscr{S}(A)_k^{-}$.
\end{enumerate}
\end{corollary}
\begin{proof}
 We first compute the action of $[A_c]_k$ in $f \circ g$:
\[
 ([A_c]_k (f \circ g)) (\boldsymbol{a}) 
= f(g(\boldsymbol{a}|_{[A_c]_k}))|_{[A_c]_k} 
= ([A_c]_k f) (g (\boldsymbol{a}|_{[A_c]_k})|_{[A_c]_k}) 
 = ([A_c]_k f) \circ ([A_c]_k g) (\boldsymbol{a}).
\]
Consequently we have
$[A_c]_k ([A_c]_k f \circ g) = f \circ [A_c]_k g$.
Hence the results follow from Proposition \ref{prop:2.1}.
\end{proof}
Among various properties of eigensequences, one of the easiest 
is \eqref{eq:3.3}:
\[
 [0] \boldsymbol{a} = 0 \text{ if } \boldsymbol{a}\in \mathscr{S}^{-}.
\]
Thus by Proposition \ref{prop:2.1} \ref{2.1.2}, \ref{2.1.4} and 
Corollary \ref{cor:2.3} the following 
equalities hold for \textit{any} sequence $\boldsymbol{a} \in \mathscr{S}$ and 
any $f,g \in \mathrm{End} (\mathscr{S})$:
\begin{align}
 &amp; [0] f(\boldsymbol{a}) = [0] ([A_c]_k f) (\boldsymbol{a}|_{[A_c]_k}), 
\label{eq:6.1} \\ 
&amp; [0] ([A_c]_k f ) (\boldsymbol{a}) = [0] f (\boldsymbol{a}|_{[A_c]_k}), 
\label{eq:6.2}\\
&amp; [0] ([A_c]_k f \circ g) (\boldsymbol{a}) = [0] (f \circ [A_c]_k g) 
(\boldsymbol{a}|_{[A_c]_k}). \label{eq:6.3}
\end{align}
For every $f$ and $g$, we obtain an identity involving $\boldsymbol{a}$.
We give several explicit examples of such formulas in the following 
proposition. We restrict ourselves again to the case $k=1$ for simplicity.

\begin{proposition} \label{prop:4.1}
 Let $\boldsymbol{a}=\{ a_n \}$  be
any sequence and $a_n^* =[n] (\boldsymbol{a}|_{[A_c]_1})$.
Then the following identities hold.

\begin{enumerate}
 \item $\displaystyle \sum_{i=0}^m \binom{m}{i} (-1)^i a_i
= \sum_{i=0}^m \binom{m}{i} (1+c)^{m-i} a_i^*$.
 \item  \label{eq:4.1.2}
$\displaystyle (-1)^n \sum_{i=0}^n \binom{n}{i} c^{n-i} a_{m+i}
=(-1)^m \sum_{i=0}^m \binom{m}{i} c^{m-i} a_{n+i}^*$.
 \item $\displaystyle \sum_{i=0}^m \binom{m}{i} (2c)^{m-i} a_{m+i} 
=\sum_{i=0}^m \binom{m}{i} (-c^2)^{m-i}a^{*}_{2i}$. \label{eq:4.1.7}
\end{enumerate}
Note that the roles of $a_n$ and $a_n^*$ are interchangeable.
\end{proposition}
 \begin{proof}
  Let $\boldsymbol{a}^* = \boldsymbol{a}|_{[A_c]_1}$.
 The following identities are used to obtain the first two 
formulas:
\begin{enumerate}
 \item $[0](\Delta^m (\boldsymbol{a})) = [0] ([A_c]_1 \Delta^m (\boldsymbol{a^*}))$ (we take $f=\Delta^m$ in \eqref{eq:6.1});
 \item $[0](L^m \circ [A_c]_1 L^n (\boldsymbol{a})) = 
 [0] ([A_c]_1 L^m \circ L^n (\boldsymbol{a}^*)) $ (we take $f=L^m$ and
$g=L^n$ in \eqref{eq:6.3}).
\end{enumerate}
Here we have already computed the explicit formulas for 
$[A_c]_1 \Delta^m$ and $[A_c]_1 L^m$ in Propositions \ref{prop:5.5.4}
and \ref{prop:5.5.5}.
The identity \ref{eq:4.1.7} is obtained by taking $f=(L^2 -c^2)^m$ and
using \eqref{eq:6.2} since we have 
$[n] ( L^2 -c^2 ) (\boldsymbol{a}) = a_{n+2} -c^2 a_n$
and $[n]([A_c]_1 L^2 -c^2 )(\boldsymbol{a}) 
= 2c a_{n+1} + a_{n+2}$.
 \end{proof}

Note that \ref{eq:4.1.2} is a generalization of the formula 
due to Chen \cite[Theorem 2.1]{MR2320995} and 
Gessel \cite[Theorem 7.4]{MR2022347}.
Kaneko's recursion formula \eqref{eq:1.2} 
is an easy consequence of \ref{eq:4.1.2} in the preceding proposition
as is noted by Gessel \cite[Lemma 7.2]{MR2022347}.
The third formula in the proposition seems to be not known before.

These identities will be simple and particularly interesting 
if $\boldsymbol{a}|_{[A_c]_k}$ is simple.
This is the case if $\boldsymbol{a}$ is an eigensequence in
$\mathscr{S}(A_c)_k^{\pm} $.

In addition to eigensequences, there are 
some interesting pairs of $\boldsymbol{a}$ and $\boldsymbol{a}|_{[A_c]_k}$:
\begin{enumerate}
 \item $\boldsymbol{a}= \{ n! \} $ and derangement numbers $D(n)$ with
       $[A_{-1}]_1$ (\cite[Section 7]{MR2022347});
 \item The special values $\xi_k (-n)$ of negative integers of
Arakawa-Kaneko zeta function and the poly-Bernoulli number $B_n^{(k)}$
with $[A_{-1}]_1$ (\cite[Theorem 6]{MR1684557}).
\end{enumerate}


As we have seen in Proposition \ref{prop:2.5} and 
Lemma \ref{lem:3.1}, the terms in
eigensequences satisfy simple relations.
These relations can be used to obtain identities for eigensequences.

\begin{proposition} \label{prop:5.2}
Assume that $c \neq 0$.
Let $\boldsymbol{a} \in \mathscr{S}(A_c)_1^{s_1}$ and 
$\boldsymbol{b} \in \mathscr{S}(A_{-c})_1^{s_2}$. Then we have
\[
[n] (\boldsymbol{a} \underset{1}{\ast} \boldsymbol{b} ) =
 \sum_{i=0}^n \binom{n}{i} \  a_{i} b_{n-i}
=0
\]
either $ s_1 s_2 = 1$ and $n$ is odd or
$ s_1 s_2 =-1$ and $i$ is even.
\end{proposition}
\begin{proof}
 By Proposition \ref{prop:2.10} we have $\boldsymbol{a}  \underset{1}{\ast}
\boldsymbol{b} \in \mathscr{S}(A_{0})_1^{s_1 s_2}$.
By Proposition \ref{prop:2.5}, the exponential generating function of 
$\boldsymbol{a}  \underset{1}{\ast}
\boldsymbol{b}$ is an even (resp. odd) power series if $s_1 s_2=1
 (\text{ resp. } s_1 s_2 =-1)$.
The proposition is now clear from this.
\end{proof}

The identities 
\[
 \begin{cases}
\displaystyle
\sum_{i=0}^n \binom{n}{i} B_i L_{n-i} =0,  &amp; \text{ if $n$ is odd}; \\
\displaystyle
\sum_{i=0}^n \binom{n}{i} B_i F_{n-i} =0,  &amp; \text{ if $n$ is even} \\
 \end{cases}
\]
involving the Lucas, Bernoulli and Fibonacci numbers proved 
in \cite[(4.1) and (4.2)]{MR1851531}
follow readily from Proposition \ref{prop:5.2}

Along with various differential operators, we can obtain more identities.

\begin{corollary} \label{prop:5.5}
 If $\boldsymbol{a} \in \mathscr{S} (A_{c})_1^{+}$ (resp. 
$\mathscr{S} (A_{c})_1^{-}$), then we have
\[
 c \sum_{j=0}^n \binom{n}{j} (-1)^{n-j} a_j a_{n-j} 
+2 \sum_{j=0}^n \binom{n}{j} (-1)^{n-j} a_j a_{n-j+1} =0
\]
for even $n$ (resp. odd $n$).
\end{corollary}

\begin{proof}
Let $\boldsymbol{a} \in \mathscr{S} (A_{c})_1^{+}$.
Then it follows from \eqref{eq:4.4} that
\[
\left( 2+ c \frac{d}{dx}\right) G_1 \left(\boldsymbol{a}, x \right) \in
\mathscr{S}_1 (A_{c})_1^{-}
\]
and by Proposition \ref{prop:2.11} we have
\[
\exp (-x) \cdot \left( 2+ c \frac{d}{dx}\right) 
G_1 \left(\boldsymbol{a}, x \right) \in
\mathscr{S}_1 (A_{-c})_1^{-}.
\]
Therefore we obtain
\[
 \left\{ G_1 (\boldsymbol{a},x), \exp (-x) \cdot \left( \frac{c}{2} +
 \frac{d}{dx}\right)
 G_1 (\boldsymbol{a},x) \right\}_0 \in \mathscr{S}_1 (A_{0})_1^{-},
\]
which is an odd power series by Proposition \ref{prop:2.5}.
The left hand side of the identity in the statement is the $n$-th term of
this sequence (see \eqref{eq:4.8}).
This proves the assertion.
The proof for $\boldsymbol{a} \in \mathscr{S} (A_{c})_1^{-}$ is
 similar.
\end{proof}

\begin{proposition} \label{prop:5.3}
 Let $\boldsymbol{a} \in \mathscr{S}(A_c)_1^{s}$.
Choose a positive integer $j$ such that $(-1)^j s =-1$.
Then we have
\begin{equation} \label{eq:5.1}
 \sum_{i=0}^{n} \binom{n}{i} \   c^{n-i} (n+i)^{\underline{j}} \ a_{n-j+i} =0.
\end{equation}
\end{proposition}
\begin{proof}
 By Example \ref{ex:4.3}, we have 
$ \{ i^{\underline{j}} \ a_{i-j} \}_{i \ge 0}
 \in \mathscr{S}(A_c)^{-}_1 $. By applying $c \partial +\partial^2$ in
\eqref{eq:4.5} $n$ times, the $m$-th term of the new sequence 
is 
\[
 \sum_{i=0}^{n} \binom{n}{i} \ c^{n-i} (n+i+m)^{\underline{j}} \ a_{n+i+m-j}.
\]
Since this sequence belongs to $\mathscr{S}(A_c)^{-}_1$, 
the $0$-th term is $0$
by Lemma \ref{lem:3.1}:
\[
 \sum_{i=0}^{n} \binom{n}{i} \ c^{n-i} (n+i)^{\underline{j}} \ a_{n+i-j}=0.
\]
\end{proof}

The following  corollary readily follows from \eqref{eq:5.1} by setting $j=1$.
\begin{corollary}[Generalized Kaneko's identity] \label{cor:5.4}
If $\boldsymbol{a} \in \mathscr{S}(A_c)_1^{+}$, then
\[
  \sum_{i=0}^{n} \binom{n}{i} \ c^{n-i} (n+i) a_{n+i-1} =0.
\]
\end{corollary}
By noting that the Bernoulli numbers $\boldsymbol{B} \in 
\mathscr{S}(A_1)_1^{+}$ (see Example \ref{ex:3.6}), 
the original formula \eqref{eq:2.2} follows by changing $n$ by $n+1$.

As we have seen, 
many identities for eigensequences can be obtained from our study on the 
involutions instead of long and complicate computations of iterated sums.

We hope that our method using the involutions and the differential operators 
shed new light on the study of sequences.

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\end{thebibliography}




\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B75; Secondary 11B37, 11B68, 05A19, 05A15. 

\noindent \emph{Keywords: } 
generating function, differential operator, linear recurrence.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received November 18 2012;
revised version received  January 16 2013.
Published in {\it Journal of Integer Sequences}, January 17 2013.

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